GCSE Maths · Sequences and nth Term · Higher · Lesson 11.8

GCSE Maths: Quadratic sequences (Higher)

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11.8

Quadratic sequences

Edexcel 1MA1: A24, A25
Grade 5-9

In an arithmetic sequence the differences between terms are all the same. Look at 3, 8, 15, 24, 35. The first differences are 5, 7, 9, 11. They are not the same, so this is not an arithmetic sequence. But they change in a steady way: each is 2 more than the one before. If we find the differences of the differences we get 2, 2, 2. These are the second differences.

Terms (dark), first differences (orange) and second differences (green). The second differences are all 2.

A sequence whose second differences are constant is a quadratic sequence. Its nth term has an n² in it and looks like an² + bn + c. Its graph is a curve, not a straight line.

The terms of n² + 2n. They lie on a curve.

Why the second difference gives a. The square numbers 1, 4, 9, 16, 25 have first differences 3, 5, 7, 9 and second differences 2, 2, 2. So the sequence n² has second difference 2. If every term is multiplied by a, the differences are all multiplied by a too, so an² has second difference 2a. Adding a linear part bn + c does not change the second differences (a linear sequence has constant first differences, so its second differences are 0). Therefore a = (second difference) ÷ 2.

The method.

Write the terms and find the first differences, then the second differences. Check the second differences are all the same.
Halve the second difference to get a.
Work out an² for each position (that is a × 1, a × 4, a × 9, …) and subtract it from each term.
What is left is an arithmetic sequence. Find its nth term, bn + c, using the common difference method from Section 11.4.
Write an² + bn + c and check it with two terms.

If a is negative the sequence curves downwards. If the second difference is odd then a is a fraction such as 12. Quadratic sequences also come from patterns where the number of dots or tiles is made from a square and a few extras.

Key ideas

Second differences constant: the sequence is quadratic, with nth term an² + bn + c.
a = (second difference) ÷ 2.
Subtract an² from every term; the leftover is linear and gives bn + c.
Always check the formula on at least two given terms.
To test whether a number is a term, set the nth term equal to it and solve for n; n must be a positive whole number.

Worked example 11.8.1

Find the nth term of the quadratic sequence 3, 8, 15, 24, 35, …

Key words. A quadratic sequence has an nth term with an n² in it, such as 2n² + 3. The first differences are the gaps between neighbouring terms. The second differences are the gaps between the first differences.

Method. Find the second differences. Half of them gives the number in front of n². Subtract that part from the terms, and the leftover is linear.

Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4 5
Term 3 8 15 24 35
First differences: 5, 7, 9, 11
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: 2, 2, 2
They are all 2, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = 2 ÷ 2 = 1
The nth term starts with n².
Subtract n² from each term. This removes the quadratic part. Work out n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4 5
n² 1 4 9 16 25
n² 1 4 9 16 25
Term 3 8 15 24 35
Term − n² 2 4 6 8 10
Find the nth term of what is left. The leftover numbers form an arithmetic sequence. Find its common difference (that is the number in front of n), then the number to add, exactly as for a linear sequence.
Leftover sequence: 2, 4, 6, 8, 10
Common difference = 4 − 2 = 2, so it starts with 2n.
Adjustment: 2 − 2 = 0
Leftover nth term = 2n
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = n² + (2n) = n² + 2n

Position n	1	2	3	4	5
Term	3	8	15	24	35

Position n	1	2	3	4	5
n²	1	4	9	16	25
n²	1	4	9	16	25
Term	3	8	15	24	35
Term − n²	2	4	6	8	10

Check. Test the formula. n = 2: 2² + 2 × 2 = 4 + 2 × 2 = 4 + 4 = 8 ✓ n = 5: 5² + 2 × 5 = 25 + 2 × 5 = 25 + 10 = 35 ✓

Watch out. a is half of the second difference, not the second difference itself. And remember to subtract an² (using each position) before finding the linear part.

Worked example 11.8.2

Find the nth term of the sequence 6, 15, 28, 45, 66, …

Method. Find the second differences. Half of them gives the number in front of n². Subtract that part from the terms, and the leftover is linear.

Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4 5
Term 6 15 28 45 66
First differences: 9, 13, 17, 21
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: 4, 4, 4
They are all 4, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = 4 ÷ 2 = 2
The nth term starts with 2n².
Subtract 2n² from each term. This removes the quadratic part. Work out 2n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4 5
n² 1 4 9 16 25
2n² 2 8 18 32 50
Term 6 15 28 45 66
Term − 2n² 4 7 10 13 16
Find the nth term of what is left. The leftover numbers form an arithmetic sequence. Find its common difference (that is the number in front of n), then the number to add, exactly as for a linear sequence.
Leftover sequence: 4, 7, 10, 13, 16
Common difference = 7 − 4 = 3, so it starts with 3n.
Adjustment: 4 − 3 = 1
Leftover nth term = 3n + 1
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = 2n² + (3n + 1) = 2n² + 3n + 1

Position n	1	2	3	4	5
Term	6	15	28	45	66

Position n	1	2	3	4	5
n²	1	4	9	16	25
2n²	2	8	18	32	50
Term	6	15	28	45	66
Term − 2n²	4	7	10	13	16

Check. Test the formula. n = 2: 2 × 2² + 3 × 2 + 1 = 2 × 4 + 3 × 2 + 1 = 8 + 6 + 1 = 15 ✓ n = 5: 2 × 5² + 3 × 5 + 1 = 2 × 25 + 3 × 5 + 1 = 50 + 15 + 1 = 66 ✓

Watch out. a is half of the second difference, not the second difference itself. And remember to subtract an² (using each position) before finding the linear part.

Worked example 11.8.3

Find the nth term of the sequence 1, 10, 25, 46, 73, …

Method. Find the second differences. Half of them gives the number in front of n². Subtract that part from the terms, and the leftover is linear.

Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4 5
Term 1 10 25 46 73
First differences: 9, 15, 21, 27
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: 6, 6, 6
They are all 6, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = 6 ÷ 2 = 3
The nth term starts with 3n².
Subtract 3n² from each term. This removes the quadratic part. Work out 3n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4 5
n² 1 4 9 16 25
3n² 3 12 27 48 75
Term 1 10 25 46 73
Term − 3n² −2 −2 −2 −2 −2
Look at what is left. The numbers left over are all the same, so there is no n term, just a number.
Every remainder is −2, so b = 0 and c = −2.
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = 3n² − 2

Position n	1	2	3	4	5
Term	1	10	25	46	73

Position n	1	2	3	4	5
n²	1	4	9	16	25
3n²	3	12	27	48	75
Term	1	10	25	46	73
Term − 3n²	−2	−2	−2	−2	−2

Check. Test the formula. n = 2: 3 × 2² − 2 = 3 × 4 − 2 = 12 − 2 = 10 ✓ n = 5: 3 × 5² − 2 = 3 × 25 − 2 = 75 − 2 = 73 ✓

Watch out. a is half of the second difference, not the second difference itself. And remember to subtract an² (using each position) before finding the linear part.

Worked example 11.8.4

Find the nth term of the sequence 12, 19, 24, 27, 28, … (the terms go up and then level off).

Method. Find the second differences. Half of them gives the number in front of n². Subtract that part from the terms, and the leftover is linear.

Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4 5
Term 12 19 24 27 28
First differences: 7, 5, 3, 1
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: −2, −2, −2
They are all −2, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = −2 ÷ 2 = −1
The nth term starts with −n².
Subtract −n² from each term. This removes the quadratic part. Work out −n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4 5
n² 1 4 9 16 25
−n² −1 −4 −9 −16 −25
Term 12 19 24 27 28
Term − −n² 13 23 33 43 53
Find the nth term of what is left. The leftover numbers form an arithmetic sequence. Find its common difference (that is the number in front of n), then the number to add, exactly as for a linear sequence.
Leftover sequence: 13, 23, 33, 43, 53
Common difference = 23 − 13 = 10, so it starts with 10n.
Adjustment: 13 − 10 = 3
Leftover nth term = 10n + 3
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = −n² + (10n + 3) = −n² + 10n + 3

Position n	1	2	3	4	5
Term	12	19	24	27	28

Position n	1	2	3	4	5
n²	1	4	9	16	25
−n²	−1	−4	−9	−16	−25
Term	12	19	24	27	28
Term − −n²	13	23	33	43	53

Check. Test the formula. n = 2: −2² + 10 × 2 + 3 = −4 + 10 × 2 + 3 = −4 + 20 + 3 = 19 ✓ n = 5: −5² + 10 × 5 + 3 = −25 + 10 × 5 + 3 = −25 + 50 + 3 = 28 ✓

Watch out. a is half of the second difference, not the second difference itself. And remember to subtract an² (using each position) before finding the linear part.

Worked example 11.8.5

Find the nth term of the sequence 1, 3, 6, 10, 15, … (the triangular numbers).

Method. Find the second differences. Half of them gives the number in front of n². Subtract that part from the terms, and the leftover is linear.

Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4 5
Term 1 3 6 10 15
First differences: 2, 3, 4, 5
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: 1, 1, 1
They are all 1, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = 1 ÷ 2 = 12
The nth term starts with 12n².
Subtract 12n² from each term. This removes the quadratic part. Work out 12n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4 5
n² 1 4 9 16 25
12n² 12 2 92 8 252
Term 1 3 6 10 15
Term − 12n² 12 1 32 2 52
Find the nth term of what is left. The leftover numbers form an arithmetic sequence. Find its common difference (that is the number in front of n), then the number to add, exactly as for a linear sequence.
Leftover sequence: 12, 1, 32, 2, 52
Common difference = 1 − 12 = 12, so it starts with 12n.
Adjustment: 12 − 12 = 0
Leftover nth term = 12n
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = 12n² + (12n) = 12n² + 12n

Position n	1	2	3	4	5
Term	1	3	6	10	15

Position n	1	2	3	4	5
n²	1	4	9	16	25
12n²	12	2	92	8	252
Term	1	3	6	10	15
Term − 12n²	12	1	32	2	52

Check. Test the formula. n = 2: 12 × 2² + 12 × 2 = 12 × 4 + 12 × 2 = 2 + 1 = 3 ✓ n = 5: 12 × 5² + 12 × 5 = 12 × 25 + 12 × 5 = 252 + 52 = 15 ✓

Watch out. a is half of the second difference, not the second difference itself. And remember to subtract an² (using each position) before finding the linear part.

Worked example 11.8.6

The nth term of a sequence is n² + 2n. Is 80 a term of the sequence?

Method. Set the nth term equal to the number and solve for n. The number is in the sequence only if n is a positive whole number.

Set the nth term equal to the number. The number is in the sequence only if some whole-number position n gives it, so we solve for n.
n² + 2n = 80
Rearrange so that one side is 0. Subtract the number from both sides. A quadratic equation is easiest to solve when one side is zero.
n² + 2n − 80 = 0
Factorise. Write it as two brackets multiplied together.
Spot the pattern. This is a quadratic of the form n² + bn + c, with b = 2 and c = -80.
We need two numbers that multiply to -80 (the last number) and add to 2 (the middle number). List the pairs that multiply to -80:
1 × -80 = -80, add: 1 + -80 = -79 ✗
-1 × 80 = -80, add: -1 + 80 = 79 ✗
2 × -40 = -80, add: 2 + -40 = -38 ✗
-2 × 40 = -80, add: -2 + 40 = 38 ✗
4 × -20 = -80, add: 4 + -20 = -16 ✗
-4 × 20 = -80, add: -4 + 20 = 16 ✗
5 × -16 = -80, add: 5 + -16 = -11 ✗
-5 × 16 = -80, add: -5 + 16 = 11 ✗
8 × -10 = -80, add: 8 + -10 = -2 ✗
-8 × 10 = -80, add: -8 + 10 = 2 ✓ this pair works
Put the pair into two brackets: (n − 8)(n + 10).
Check by expanding: n² + 10n − 8n − 80 = n² + 2n − 80. ✓
Solve: one bracket must equal 0. If two things multiply to give 0 then at least one of them is 0.
n − 8 = 0 gives n = 8
n + 10 = 0 gives n = −10
Decide. A position must be a whole number that is at least 1. Ignore a negative answer.
Yes: 80 is the 8th term (the other answer is not a position).

Check. Substitute n = 8: 8² + 2 × 8 = 64 + 2 × 8 = 64 + 16 = 80. ✓

Watch out. A quadratic equation can have two answers. Keep only a positive whole number as the position.

Worked example 11.8.7

Here are the first four patterns of dots.

Find the nth term, and the number of dots in pattern 20.

Key words. The position of a term is its place in the sequence: 1st, 2nd, 3rd and so on. We use the letter n for the position number. The nth term (or position-to-term rule) is a formula that gives the term for any position n.

Method. Count the dots, find the rule for the sequence of counts, and use the picture to see why it works.

Count the dots in each pattern. Count carefully, one pattern at a time. This turns the pictures into a sequence of numbers.
Pattern number n 1 2 3 4
Number of dots 3 8 15 24
Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4
Term 3 8 15 24
First differences: 5, 7, 9
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: 2, 2
They are all 2, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = 2 ÷ 2 = 1
The nth term starts with n².
Subtract n² from each term. This removes the quadratic part. Work out n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4
n² 1 4 9 16
n² 1 4 9 16
Term 3 8 15 24
Term − n² 2 4 6 8
Find the nth term of what is left. The leftover numbers form an arithmetic sequence. Find its common difference (that is the number in front of n), then the number to add, exactly as for a linear sequence.
Leftover sequence: 2, 4, 6, 8
Common difference = 4 − 2 = 2, so it starts with 2n.
Adjustment: 2 − 2 = 0
Leftover nth term = 2n
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = n² + (2n) = n² + 2n
Check that the rule makes sense with the picture. A good formula can be explained from the drawing.
Pattern n is a rectangle of dots with n rows and n + 2 columns, so it has n(n + 2) = n² + 2n dots.
Replace n by 20. Pattern 20 is the one in position 20.
Pattern 20: 20² + 2 × 20 = 400 + 2 × 20 = 400 + 40 = 440

Pattern number n	1	2	3	4
Number of dots	3	8	15	24

Position n	1	2	3	4
Term	3	8	15	24

Position n	1	2	3	4
n²	1	4	9	16
n²	1	4	9	16
Term	3	8	15	24
Term − n²	2	4	6	8

Check. Test the rule on patterns you can see. n = 1: 1² + 2 × 1 = 1 + 2 × 1 = 1 + 2 = 3 ✓ n = 4: 4² + 2 × 4 = 16 + 2 × 4 = 16 + 8 = 24 ✓

Watch out. Do not just keep adding on a drawing for a large pattern number. Find the nth term and substitute.

Watch out

a is half of the second difference, not the second difference itself. In 3, 8, 15, 24 the second difference is 2 so a = 1, not 2. Also remember to subtract an² using the position of each term: 1², 2², 3², … (that is 1, 4, 9, …).

Exercise 11H Quadratic sequences

Non-calculator

FFind the nth term of each quadratic sequence.

(a)3, 8, 15, 24, 35, …

Method. Find the second differences. Half of them gives the number in front of n². Subtract that part from the terms, and the leftover is linear.

Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4 5
Term 3 8 15 24 35
First differences: 5, 7, 9, 11
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: 2, 2, 2
They are all 2, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = 2 ÷ 2 = 1
The nth term starts with n².
Subtract n² from each term. This removes the quadratic part. Work out n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4 5
n² 1 4 9 16 25
n² 1 4 9 16 25
Term 3 8 15 24 35
Term − n² 2 4 6 8 10
Find the nth term of what is left. The leftover numbers form an arithmetic sequence. Find its common difference (that is the number in front of n), then the number to add, exactly as for a linear sequence.
Leftover sequence: 2, 4, 6, 8, 10
Common difference = 4 − 2 = 2, so it starts with 2n.
Adjustment: 2 − 2 = 0
Leftover nth term = 2n
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = n² + (2n) = n² + 2n

Position n	1	2	3	4	5
Term	3	8	15	24	35

Position n	1	2	3	4	5
n²	1	4	9	16	25
n²	1	4	9	16	25
Term	3	8	15	24	35
Term − n²	2	4	6	8	10

Check. Test the formula. n = 2: 2² + 2 × 2 = 4 + 2 × 2 = 4 + 4 = 8 ✓ n = 5: 5² + 2 × 5 = 25 + 2 × 5 = 25 + 10 = 35 ✓

Watch out. a is half of the second difference, not the second difference itself. And remember to subtract an² (using each position) before finding the linear part.

(b)5, 8, 13, 20, 29, …

Method. Find the second differences. Half of them gives the number in front of n². Subtract that part from the terms, and the leftover is linear.

Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4 5
Term 5 8 13 20 29
First differences: 3, 5, 7, 9
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: 2, 2, 2
They are all 2, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = 2 ÷ 2 = 1
The nth term starts with n².
Subtract n² from each term. This removes the quadratic part. Work out n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4 5
n² 1 4 9 16 25
n² 1 4 9 16 25
Term 5 8 13 20 29
Term − n² 4 4 4 4 4
Look at what is left. The numbers left over are all the same, so there is no n term, just a number.
Every remainder is 4, so b = 0 and c = 4.
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = n² + 4

Position n	1	2	3	4	5
Term	5	8	13	20	29

Position n	1	2	3	4	5
n²	1	4	9	16	25
n²	1	4	9	16	25
Term	5	8	13	20	29
Term − n²	4	4	4	4	4

Check. Test the formula. n = 2: 2² + 4 = 4 + 4 = 8 ✓ n = 5: 5² + 4 = 25 + 4 = 29 ✓

Watch out. a is half of the second difference, not the second difference itself. And remember to subtract an² (using each position) before finding the linear part.

Method. Find the second differences. Half of them gives the number in front of n². Subtract that part from the terms, and the leftover is linear.

Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4 5
Term 2 8 16 26 38
First differences: 6, 8, 10, 12
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: 2, 2, 2
They are all 2, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = 2 ÷ 2 = 1
The nth term starts with n².
Subtract n² from each term. This removes the quadratic part. Work out n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4 5
n² 1 4 9 16 25
n² 1 4 9 16 25
Term 2 8 16 26 38
Term − n² 1 4 7 10 13
Find the nth term of what is left. The leftover numbers form an arithmetic sequence. Find its common difference (that is the number in front of n), then the number to add, exactly as for a linear sequence.
Leftover sequence: 1, 4, 7, 10, 13
Common difference = 4 − 1 = 3, so it starts with 3n.
Adjustment: 1 − 3 = −2
Leftover nth term = 3n − 2
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = n² + (3n − 2) = n² + 3n − 2

Position n	1	2	3	4	5
Term	2	8	16	26	38

Position n	1	2	3	4	5
n²	1	4	9	16	25
n²	1	4	9	16	25
Term	2	8	16	26	38
Term − n²	1	4	7	10	13

Check. Test the formula. n = 2: 2² + 3 × 2 − 2 = 4 + 3 × 2 − 2 = 4 + 6 − 2 = 8 ✓ n = 5: 5² + 3 × 5 − 2 = 25 + 3 × 5 − 2 = 25 + 15 − 2 = 38 ✓

Watch out. a is half of the second difference, not the second difference itself. And remember to subtract an² (using each position) before finding the linear part.

(d)5, 7, 11, 17, 25, …

Method. Find the second differences. Half of them gives the number in front of n². Subtract that part from the terms, and the leftover is linear.

Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4 5
Term 5 7 11 17 25
First differences: 2, 4, 6, 8
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: 2, 2, 2
They are all 2, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = 2 ÷ 2 = 1
The nth term starts with n².
Subtract n² from each term. This removes the quadratic part. Work out n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4 5
n² 1 4 9 16 25
n² 1 4 9 16 25
Term 5 7 11 17 25
Term − n² 4 3 2 1 0
Find the nth term of what is left. The leftover numbers form an arithmetic sequence. Find its common difference (that is the number in front of n), then the number to add, exactly as for a linear sequence.
Leftover sequence: 4, 3, 2, 1, 0
Common difference = 3 − 4 = −1, so it starts with −n.
Adjustment: 4 − (−1) = 5
Leftover nth term = −n + 5
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = n² + (−n + 5) = n² − n + 5

Position n	1	2	3	4	5
Term	5	7	11	17	25

Position n	1	2	3	4	5
n²	1	4	9	16	25
n²	1	4	9	16	25
Term	5	7	11	17	25
Term − n²	4	3	2	1	0

Check. Test the formula. n = 2: 2² − 2 + 5 = 4 − 2 + 5 = 7 ✓ n = 5: 5² − 5 + 5 = 25 − 5 + 5 = 25 ✓

Watch out. a is half of the second difference, not the second difference itself. And remember to subtract an² (using each position) before finding the linear part.

(e)6, 13, 22, 33, 46, …

Method. Find the second differences. Half of them gives the number in front of n². Subtract that part from the terms, and the leftover is linear.

Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4 5
Term 6 13 22 33 46
First differences: 7, 9, 11, 13
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: 2, 2, 2
They are all 2, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = 2 ÷ 2 = 1
The nth term starts with n².
Subtract n² from each term. This removes the quadratic part. Work out n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4 5
n² 1 4 9 16 25
n² 1 4 9 16 25
Term 6 13 22 33 46
Term − n² 5 9 13 17 21
Find the nth term of what is left. The leftover numbers form an arithmetic sequence. Find its common difference (that is the number in front of n), then the number to add, exactly as for a linear sequence.
Leftover sequence: 5, 9, 13, 17, 21
Common difference = 9 − 5 = 4, so it starts with 4n.
Adjustment: 5 − 4 = 1
Leftover nth term = 4n + 1
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = n² + (4n + 1) = n² + 4n + 1

Position n	1	2	3	4	5
Term	6	13	22	33	46

Position n	1	2	3	4	5
n²	1	4	9	16	25
n²	1	4	9	16	25
Term	6	13	22	33	46
Term − n²	5	9	13	17	21

Check. Test the formula. n = 2: 2² + 4 × 2 + 1 = 4 + 4 × 2 + 1 = 4 + 8 + 1 = 13 ✓ n = 5: 5² + 4 × 5 + 1 = 25 + 4 × 5 + 1 = 25 + 20 + 1 = 46 ✓

Watch out. a is half of the second difference, not the second difference itself. And remember to subtract an² (using each position) before finding the linear part.

(f)0, 3, 8, 15, 24, …

Method. Find the second differences. Half of them gives the number in front of n². Subtract that part from the terms, and the leftover is linear.

Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4 5
Term 0 3 8 15 24
First differences: 3, 5, 7, 9
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: 2, 2, 2
They are all 2, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = 2 ÷ 2 = 1
The nth term starts with n².
Subtract n² from each term. This removes the quadratic part. Work out n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4 5
n² 1 4 9 16 25
n² 1 4 9 16 25
Term 0 3 8 15 24
Term − n² −1 −1 −1 −1 −1
Look at what is left. The numbers left over are all the same, so there is no n term, just a number.
Every remainder is −1, so b = 0 and c = −1.
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = n² − 1

Position n	1	2	3	4	5
Term	0	3	8	15	24

Position n	1	2	3	4	5
n²	1	4	9	16	25
n²	1	4	9	16	25
Term	0	3	8	15	24
Term − n²	−1	−1	−1	−1	−1

Check. Test the formula. n = 2: 2² − 1 = 4 − 1 = 3 ✓ n = 5: 5² − 1 = 25 − 1 = 24 ✓

Watch out. a is half of the second difference, not the second difference itself. And remember to subtract an² (using each position) before finding the linear part.

Done

FFind the nth term of each quadratic sequence.

(a)6, 15, 28, 45, 66, …

Method. Find the second differences. Half of them gives the number in front of n². Subtract that part from the terms, and the leftover is linear.

Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4 5
Term 6 15 28 45 66
First differences: 9, 13, 17, 21
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: 4, 4, 4
They are all 4, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = 4 ÷ 2 = 2
The nth term starts with 2n².
Subtract 2n² from each term. This removes the quadratic part. Work out 2n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4 5
n² 1 4 9 16 25
2n² 2 8 18 32 50
Term 6 15 28 45 66
Term − 2n² 4 7 10 13 16
Find the nth term of what is left. The leftover numbers form an arithmetic sequence. Find its common difference (that is the number in front of n), then the number to add, exactly as for a linear sequence.
Leftover sequence: 4, 7, 10, 13, 16
Common difference = 7 − 4 = 3, so it starts with 3n.
Adjustment: 4 − 3 = 1
Leftover nth term = 3n + 1
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = 2n² + (3n + 1) = 2n² + 3n + 1

Position n	1	2	3	4	5
Term	6	15	28	45	66

Position n	1	2	3	4	5
n²	1	4	9	16	25
2n²	2	8	18	32	50
Term	6	15	28	45	66
Term − 2n²	4	7	10	13	16

Check. Test the formula. n = 2: 2 × 2² + 3 × 2 + 1 = 2 × 4 + 3 × 2 + 1 = 8 + 6 + 1 = 15 ✓ n = 5: 2 × 5² + 3 × 5 + 1 = 2 × 25 + 3 × 5 + 1 = 50 + 15 + 1 = 66 ✓

Watch out. a is half of the second difference, not the second difference itself. And remember to subtract an² (using each position) before finding the linear part.

(b)6, 14, 28, 48, 74, …

Method. Find the second differences. Half of them gives the number in front of n². Subtract that part from the terms, and the leftover is linear.

Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4 5
Term 6 14 28 48 74
First differences: 8, 14, 20, 26
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: 6, 6, 6
They are all 6, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = 6 ÷ 2 = 3
The nth term starts with 3n².
Subtract 3n² from each term. This removes the quadratic part. Work out 3n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4 5
n² 1 4 9 16 25
3n² 3 12 27 48 75
Term 6 14 28 48 74
Term − 3n² 3 2 1 0 −1
Find the nth term of what is left. The leftover numbers form an arithmetic sequence. Find its common difference (that is the number in front of n), then the number to add, exactly as for a linear sequence.
Leftover sequence: 3, 2, 1, 0, −1
Common difference = 2 − 3 = −1, so it starts with −n.
Adjustment: 3 − (−1) = 4
Leftover nth term = −n + 4
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = 3n² + (−n + 4) = 3n² − n + 4

Position n	1	2	3	4	5
Term	6	14	28	48	74

Position n	1	2	3	4	5
n²	1	4	9	16	25
3n²	3	12	27	48	75
Term	6	14	28	48	74
Term − 3n²	3	2	1	0	−1

Check. Test the formula. n = 2: 3 × 2² − 2 + 4 = 3 × 4 − 2 + 4 = 12 − 2 + 4 = 14 ✓ n = 5: 3 × 5² − 5 + 4 = 3 × 25 − 5 + 4 = 75 − 5 + 4 = 74 ✓

Watch out. a is half of the second difference, not the second difference itself. And remember to subtract an² (using each position) before finding the linear part.

Method. Find the second differences. Half of them gives the number in front of n². Subtract that part from the terms, and the leftover is linear.

Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4 5
Term 1 13 33 61 97
First differences: 12, 20, 28, 36
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: 8, 8, 8
They are all 8, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = 8 ÷ 2 = 4
The nth term starts with 4n².
Subtract 4n² from each term. This removes the quadratic part. Work out 4n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4 5
n² 1 4 9 16 25
4n² 4 16 36 64 100
Term 1 13 33 61 97
Term − 4n² −3 −3 −3 −3 −3
Look at what is left. The numbers left over are all the same, so there is no n term, just a number.
Every remainder is −3, so b = 0 and c = −3.
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = 4n² − 3

Position n	1	2	3	4	5
Term	1	13	33	61	97

Position n	1	2	3	4	5
n²	1	4	9	16	25
4n²	4	16	36	64	100
Term	1	13	33	61	97
Term − 4n²	−3	−3	−3	−3	−3

Check. Test the formula. n = 2: 4 × 2² − 3 = 4 × 4 − 3 = 16 − 3 = 13 ✓ n = 5: 4 × 5² − 3 = 4 × 25 − 3 = 100 − 3 = 97 ✓

Watch out. a is half of the second difference, not the second difference itself. And remember to subtract an² (using each position) before finding the linear part.

(d)7, 18, 33, 52, 75, …

Method. Find the second differences. Half of them gives the number in front of n². Subtract that part from the terms, and the leftover is linear.

Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4 5
Term 7 18 33 52 75
First differences: 11, 15, 19, 23
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: 4, 4, 4
They are all 4, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = 4 ÷ 2 = 2
The nth term starts with 2n².
Subtract 2n² from each term. This removes the quadratic part. Work out 2n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4 5
n² 1 4 9 16 25
2n² 2 8 18 32 50
Term 7 18 33 52 75
Term − 2n² 5 10 15 20 25
Find the nth term of what is left. The leftover numbers form an arithmetic sequence. Find its common difference (that is the number in front of n), then the number to add, exactly as for a linear sequence.
Leftover sequence: 5, 10, 15, 20, 25
Common difference = 10 − 5 = 5, so it starts with 5n.
Adjustment: 5 − 5 = 0
Leftover nth term = 5n
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = 2n² + (5n) = 2n² + 5n

Position n	1	2	3	4	5
Term	7	18	33	52	75

Position n	1	2	3	4	5
n²	1	4	9	16	25
2n²	2	8	18	32	50
Term	7	18	33	52	75
Term − 2n²	5	10	15	20	25

Check. Test the formula. n = 2: 2 × 2² + 5 × 2 = 2 × 4 + 5 × 2 = 8 + 10 = 18 ✓ n = 5: 2 × 5² + 5 × 5 = 2 × 25 + 5 × 5 = 50 + 25 = 75 ✓

Watch out. a is half of the second difference, not the second difference itself. And remember to subtract an² (using each position) before finding the linear part.

(e)6, 23, 50, 87, 134, …

Method. Find the second differences. Half of them gives the number in front of n². Subtract that part from the terms, and the leftover is linear.

Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4 5
Term 6 23 50 87 134
First differences: 17, 27, 37, 47
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: 10, 10, 10
They are all 10, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = 10 ÷ 2 = 5
The nth term starts with 5n².
Subtract 5n² from each term. This removes the quadratic part. Work out 5n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4 5
n² 1 4 9 16 25
5n² 5 20 45 80 125
Term 6 23 50 87 134
Term − 5n² 1 3 5 7 9
Find the nth term of what is left. The leftover numbers form an arithmetic sequence. Find its common difference (that is the number in front of n), then the number to add, exactly as for a linear sequence.
Leftover sequence: 1, 3, 5, 7, 9
Common difference = 3 − 1 = 2, so it starts with 2n.
Adjustment: 1 − 2 = −1
Leftover nth term = 2n − 1
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = 5n² + (2n − 1) = 5n² + 2n − 1

Position n	1	2	3	4	5
Term	6	23	50	87	134

Position n	1	2	3	4	5
n²	1	4	9	16	25
5n²	5	20	45	80	125
Term	6	23	50	87	134
Term − 5n²	1	3	5	7	9

Check. Test the formula. n = 2: 5 × 2² + 2 × 2 − 1 = 5 × 4 + 2 × 2 − 1 = 20 + 4 − 1 = 23 ✓ n = 5: 5 × 5² + 2 × 5 − 1 = 5 × 25 + 2 × 5 − 1 = 125 + 10 − 1 = 134 ✓

Watch out. a is half of the second difference, not the second difference itself. And remember to subtract an² (using each position) before finding the linear part.

(f)4, 13, 28, 49, 76, …

Method. Find the second differences. Half of them gives the number in front of n². Subtract that part from the terms, and the leftover is linear.

Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4 5
Term 4 13 28 49 76
First differences: 9, 15, 21, 27
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: 6, 6, 6
They are all 6, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = 6 ÷ 2 = 3
The nth term starts with 3n².
Subtract 3n² from each term. This removes the quadratic part. Work out 3n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4 5
n² 1 4 9 16 25
3n² 3 12 27 48 75
Term 4 13 28 49 76
Term − 3n² 1 1 1 1 1
Look at what is left. The numbers left over are all the same, so there is no n term, just a number.
Every remainder is 1, so b = 0 and c = 1.
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = 3n² + 1

Position n	1	2	3	4	5
Term	4	13	28	49	76

Position n	1	2	3	4	5
n²	1	4	9	16	25
3n²	3	12	27	48	75
Term	4	13	28	49	76
Term − 3n²	1	1	1	1	1

Check. Test the formula. n = 2: 3 × 2² + 1 = 3 × 4 + 1 = 12 + 1 = 13 ✓ n = 5: 3 × 5² + 1 = 3 × 25 + 1 = 75 + 1 = 76 ✓

Watch out. a is half of the second difference, not the second difference itself. And remember to subtract an² (using each position) before finding the linear part.

Done

FFind the nth term of each sequence. The second difference is negative, or a is a fraction.

(a)12, 19, 24, 27, 28, …

Method. Find the second differences. Half of them gives the number in front of n². Subtract that part from the terms, and the leftover is linear.

Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4 5
Term 12 19 24 27 28
First differences: 7, 5, 3, 1
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: −2, −2, −2
They are all −2, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = −2 ÷ 2 = −1
The nth term starts with −n².
Subtract −n² from each term. This removes the quadratic part. Work out −n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4 5
n² 1 4 9 16 25
−n² −1 −4 −9 −16 −25
Term 12 19 24 27 28
Term − −n² 13 23 33 43 53
Find the nth term of what is left. The leftover numbers form an arithmetic sequence. Find its common difference (that is the number in front of n), then the number to add, exactly as for a linear sequence.
Leftover sequence: 13, 23, 33, 43, 53
Common difference = 23 − 13 = 10, so it starts with 10n.
Adjustment: 13 − 10 = 3
Leftover nth term = 10n + 3
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = −n² + (10n + 3) = −n² + 10n + 3

Position n	1	2	3	4	5
Term	12	19	24	27	28

Position n	1	2	3	4	5
n²	1	4	9	16	25
−n²	−1	−4	−9	−16	−25
Term	12	19	24	27	28
Term − −n²	13	23	33	43	53

Check. Test the formula. n = 2: −2² + 10 × 2 + 3 = −4 + 10 × 2 + 3 = −4 + 20 + 3 = 19 ✓ n = 5: −5² + 10 × 5 + 3 = −25 + 10 × 5 + 3 = −25 + 50 + 3 = 28 ✓

Watch out. a is half of the second difference, not the second difference itself. And remember to subtract an² (using each position) before finding the linear part.

(b)18, 12, 2, −12, −30, …

Method. Find the second differences. Half of them gives the number in front of n². Subtract that part from the terms, and the leftover is linear.

Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4 5
Term 18 12 2 −12 −30
First differences: −6, −10, −14, −18
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: −4, −4, −4
They are all −4, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = −4 ÷ 2 = −2
The nth term starts with −2n².
Subtract −2n² from each term. This removes the quadratic part. Work out −2n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4 5
n² 1 4 9 16 25
−2n² −2 −8 −18 −32 −50
Term 18 12 2 −12 −30
Term − −2n² 20 20 20 20 20
Look at what is left. The numbers left over are all the same, so there is no n term, just a number.
Every remainder is 20, so b = 0 and c = 20.
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = −2n² + 20

Position n	1	2	3	4	5
Term	18	12	2	−12	−30

Position n	1	2	3	4	5
n²	1	4	9	16	25
−2n²	−2	−8	−18	−32	−50
Term	18	12	2	−12	−30
Term − −2n²	20	20	20	20	20

Check. Test the formula. n = 2: −2 × 2² + 20 = −2 × 4 + 20 = −8 + 20 = 12 ✓ n = 5: −2 × 5² + 20 = −2 × 25 + 20 = −50 + 20 = −30 ✓

Watch out. a is half of the second difference, not the second difference itself. And remember to subtract an² (using each position) before finding the linear part.

Method. Find the second differences. Half of them gives the number in front of n². Subtract that part from the terms, and the leftover is linear.

Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4 5
Term 1 3 6 10 15
First differences: 2, 3, 4, 5
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: 1, 1, 1
They are all 1, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = 1 ÷ 2 = 12
The nth term starts with 12n².
Subtract 12n² from each term. This removes the quadratic part. Work out 12n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4 5
n² 1 4 9 16 25
12n² 12 2 92 8 252
Term 1 3 6 10 15
Term − 12n² 12 1 32 2 52
Find the nth term of what is left. The leftover numbers form an arithmetic sequence. Find its common difference (that is the number in front of n), then the number to add, exactly as for a linear sequence.
Leftover sequence: 12, 1, 32, 2, 52
Common difference = 1 − 12 = 12, so it starts with 12n.
Adjustment: 12 − 12 = 0
Leftover nth term = 12n
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = 12n² + (12n) = 12n² + 12n

Position n	1	2	3	4	5
Term	1	3	6	10	15

Position n	1	2	3	4	5
n²	1	4	9	16	25
12n²	12	2	92	8	252
Term	1	3	6	10	15
Term − 12n²	12	1	32	2	52

Check. Test the formula. n = 2: 12 × 2² + 12 × 2 = 12 × 4 + 12 × 2 = 2 + 1 = 3 ✓ n = 5: 12 × 5² + 12 × 5 = 12 × 25 + 12 × 5 = 252 + 52 = 15 ✓

Watch out. a is half of the second difference, not the second difference itself. And remember to subtract an² (using each position) before finding the linear part.

(d)3, 6, 10, 15, 21, …

Method. Find the second differences. Half of them gives the number in front of n². Subtract that part from the terms, and the leftover is linear.

Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4 5
Term 3 6 10 15 21
First differences: 3, 4, 5, 6
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: 1, 1, 1
They are all 1, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = 1 ÷ 2 = 12
The nth term starts with 12n².
Subtract 12n² from each term. This removes the quadratic part. Work out 12n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4 5
n² 1 4 9 16 25
12n² 12 2 92 8 252
Term 3 6 10 15 21
Term − 12n² 52 4 112 7 172
Find the nth term of what is left. The leftover numbers form an arithmetic sequence. Find its common difference (that is the number in front of n), then the number to add, exactly as for a linear sequence.
Leftover sequence: 52, 4, 112, 7, 172
Common difference = 4 − 52 = 32, so it starts with 32n.
Adjustment: 52 − 32 = 1
Leftover nth term = 32n + 1
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = 12n² + (32n + 1) = 12n² + 32n + 1

Position n	1	2	3	4	5
Term	3	6	10	15	21

Position n	1	2	3	4	5
n²	1	4	9	16	25
12n²	12	2	92	8	252
Term	3	6	10	15	21
Term − 12n²	52	4	112	7	172

Check. Test the formula. n = 2: 12 × 2² + 32 × 2 + 1 = 12 × 4 + 32 × 2 + 1 = 2 + 3 + 1 = 6 ✓ n = 5: 12 × 5² + 32 × 5 + 1 = 12 × 25 + 32 × 5 + 1 = 252 + 152 + 1 = 21 ✓

Watch out. a is half of the second difference, not the second difference itself. And remember to subtract an² (using each position) before finding the linear part.

(e)13, 18, 17, 10, −3, …

Method. Find the second differences. Half of them gives the number in front of n². Subtract that part from the terms, and the leftover is linear.

Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4 5
Term 13 18 17 10 −3
First differences: 5, −1, −7, −13
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: −6, −6, −6
They are all −6, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = −6 ÷ 2 = −3
The nth term starts with −3n².
Subtract −3n² from each term. This removes the quadratic part. Work out −3n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4 5
n² 1 4 9 16 25
−3n² −3 −12 −27 −48 −75
Term 13 18 17 10 −3
Term − −3n² 16 30 44 58 72
Find the nth term of what is left. The leftover numbers form an arithmetic sequence. Find its common difference (that is the number in front of n), then the number to add, exactly as for a linear sequence.
Leftover sequence: 16, 30, 44, 58, 72
Common difference = 30 − 16 = 14, so it starts with 14n.
Adjustment: 16 − 14 = 2
Leftover nth term = 14n + 2
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = −3n² + (14n + 2) = −3n² + 14n + 2

Position n	1	2	3	4	5
Term	13	18	17	10	−3

Position n	1	2	3	4	5
n²	1	4	9	16	25
−3n²	−3	−12	−27	−48	−75
Term	13	18	17	10	−3
Term − −3n²	16	30	44	58	72

Check. Test the formula. n = 2: −3 × 2² + 14 × 2 + 2 = −3 × 4 + 14 × 2 + 2 = −12 + 28 + 2 = 18 ✓ n = 5: −3 × 5² + 14 × 5 + 2 = −3 × 25 + 14 × 5 + 2 = −75 + 70 + 2 = −3 ✓

Watch out. a is half of the second difference, not the second difference itself. And remember to subtract an² (using each position) before finding the linear part.

Done

FHere are the first four patterns of dots.

(a)Find the nth term for the number of dots.

Method. Count the dots, find the rule for the sequence of counts, and use the picture to see why it works.

Count the dots in each pattern. Count carefully, one pattern at a time. This turns the pictures into a sequence of numbers.
Pattern number n 1 2 3 4
Number of dots 2 6 12 20
Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4
Term 2 6 12 20
First differences: 4, 6, 8
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: 2, 2
They are all 2, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = 2 ÷ 2 = 1
The nth term starts with n².
Subtract n² from each term. This removes the quadratic part. Work out n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4
n² 1 4 9 16
n² 1 4 9 16
Term 2 6 12 20
Term − n² 1 2 3 4
Find the nth term of what is left. The leftover numbers form an arithmetic sequence. Find its common difference (that is the number in front of n), then the number to add, exactly as for a linear sequence.
Leftover sequence: 1, 2, 3, 4
Common difference = 2 − 1 = 1, so it starts with n.
Adjustment: 1 − 1 = 0
Leftover nth term = n
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = n² + (n) = n² + n
Check that the rule makes sense with the picture. A good formula can be explained from the drawing.
Pattern n is a rectangle of dots with n rows and n + 1 columns, so it has n(n + 1) = n² + n dots.

Pattern number n	1	2	3	4
Number of dots	2	6	12	20

Position n	1	2	3	4
Term	2	6	12	20

Position n	1	2	3	4
n²	1	4	9	16
n²	1	4	9	16
Term	2	6	12	20
Term − n²	1	2	3	4

Check. Test the rule on patterns you can see. n = 1: 1² + 1 = 1 + 1 = 2 ✓ n = 4: 4² + 4 = 16 + 4 = 20 ✓

Watch out. Do not just keep adding on a drawing for a large pattern number. Find the nth term and substitute.

(b)How many dots are in pattern 15?

Method. Count the dots, find the rule for the sequence of counts, and use the picture to see why it works.

Count the dots in each pattern. Count carefully, one pattern at a time. This turns the pictures into a sequence of numbers.
Pattern number n 1 2 3 4
Number of dots 2 6 12 20
Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4
Term 2 6 12 20
First differences: 4, 6, 8
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: 2, 2
They are all 2, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = 2 ÷ 2 = 1
The nth term starts with n².
Subtract n² from each term. This removes the quadratic part. Work out n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4
n² 1 4 9 16
n² 1 4 9 16
Term 2 6 12 20
Term − n² 1 2 3 4
Find the nth term of what is left. The leftover numbers form an arithmetic sequence. Find its common difference (that is the number in front of n), then the number to add, exactly as for a linear sequence.
Leftover sequence: 1, 2, 3, 4
Common difference = 2 − 1 = 1, so it starts with n.
Adjustment: 1 − 1 = 0
Leftover nth term = n
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = n² + (n) = n² + n
Check that the rule makes sense with the picture. A good formula can be explained from the drawing.
Pattern n is a rectangle of dots with n rows and n + 1 columns, so it has n(n + 1) = n² + n dots.
Replace n by 15. Pattern 15 is the one in position 15.
Pattern 15: 15² + 15 = 225 + 15 = 240

Check. Test the rule on patterns you can see. n = 1: 1² + 1 = 1 + 1 = 2 ✓ n = 4: 4² + 4 = 16 + 4 = 20 ✓

Watch out. Do not just keep adding on a drawing for a large pattern number. Find the nth term and substitute.

Done

FHere are the first four patterns of dots.

(a)Find the nth term.

Method. Count the dots, find the rule for the sequence of counts, and use the picture to see why it works.

Count the dots in each pattern. Count carefully, one pattern at a time. This turns the pictures into a sequence of numbers.
Pattern number n 1 2 3 4
Number of dots 3 8 15 24
Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4
Term 3 8 15 24
First differences: 5, 7, 9
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: 2, 2
They are all 2, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = 2 ÷ 2 = 1
The nth term starts with n².
Subtract n² from each term. This removes the quadratic part. Work out n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4
n² 1 4 9 16
n² 1 4 9 16
Term 3 8 15 24
Term − n² 2 4 6 8
Find the nth term of what is left. The leftover numbers form an arithmetic sequence. Find its common difference (that is the number in front of n), then the number to add, exactly as for a linear sequence.
Leftover sequence: 2, 4, 6, 8
Common difference = 4 − 2 = 2, so it starts with 2n.
Adjustment: 2 − 2 = 0
Leftover nth term = 2n
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = n² + (2n) = n² + 2n
Check that the rule makes sense with the picture. A good formula can be explained from the drawing.
Pattern n is a rectangle of dots with n rows and n + 2 columns, so it has n(n + 2) = n² + 2n dots.

Check. Test the rule on patterns you can see. n = 1: 1² + 2 × 1 = 1 + 2 × 1 = 1 + 2 = 3 ✓ n = 4: 4² + 2 × 4 = 16 + 2 × 4 = 16 + 8 = 24 ✓

Watch out. Do not just keep adding on a drawing for a large pattern number. Find the nth term and substitute.

(b)Which pattern has exactly 80 dots?

Method. Count the dots, find the rule for the sequence of counts, and use the picture to see why it works.

Count the dots in each pattern. Count carefully, one pattern at a time. This turns the pictures into a sequence of numbers.
Pattern number n 1 2 3 4
Number of dots 3 8 15 24
Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4
Term 3 8 15 24
First differences: 5, 7, 9
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: 2, 2
They are all 2, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = 2 ÷ 2 = 1
The nth term starts with n².
Subtract n² from each term. This removes the quadratic part. Work out n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4
n² 1 4 9 16
n² 1 4 9 16
Term 3 8 15 24
Term − n² 2 4 6 8
Find the nth term of what is left. The leftover numbers form an arithmetic sequence. Find its common difference (that is the number in front of n), then the number to add, exactly as for a linear sequence.
Leftover sequence: 2, 4, 6, 8
Common difference = 4 − 2 = 2, so it starts with 2n.
Adjustment: 2 − 2 = 0
Leftover nth term = 2n
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = n² + (2n) = n² + 2n
Check that the rule makes sense with the picture. A good formula can be explained from the drawing.
Pattern n is a rectangle of dots with n rows and n + 2 columns, so it has n(n + 2) = n² + 2n dots.
Set the rule equal to the number. We want the pattern that has this many.
n² + 2n = 80
Solve. Try whole numbers, or rearrange and factorise.
n = 8, because 8² + 2 × 8 = 64 + 2 × 8 = 64 + 16 = 80
Write the answer. The pattern number is the value of n.
Pattern 8

Check. Test the rule on patterns you can see. n = 1: 1² + 2 × 1 = 1 + 2 × 1 = 1 + 2 = 3 ✓ n = 4: 4² + 2 × 4 = 16 + 2 × 4 = 16 + 8 = 24 ✓

Watch out. Do not just keep adding on a drawing for a large pattern number. Find the nth term and substitute.

Done

PSA square pond (blue) is surrounded by a path (orange). The pictures show the first three patterns. The total number of tiles, blue and orange together, is counted.

(a)Find the nth term for the total number of tiles.

Method. Count the tiles (blue and orange together), find the rule for the sequence of counts, and use the picture to see why it works.

Count the tiles (blue and orange together) in each pattern. Count carefully, one pattern at a time. This turns the pictures into a sequence of numbers.
Pattern number n 1 2 3
Number of tiles (blue and orange together) 9 16 25
Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3
Term 9 16 25
First differences: 7, 9
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: 2
They are all 2, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = 2 ÷ 2 = 1
The nth term starts with n².
Subtract n² from each term. This removes the quadratic part. Work out n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3
n² 1 4 9
n² 1 4 9
Term 9 16 25
Term − n² 8 12 16
Find the nth term of what is left. The leftover numbers form an arithmetic sequence. Find its common difference (that is the number in front of n), then the number to add, exactly as for a linear sequence.
Leftover sequence: 8, 12, 16
Common difference = 12 − 8 = 4, so it starts with 4n.
Adjustment: 8 − 4 = 4
Leftover nth term = 4n + 4
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = n² + (4n + 4) = n² + 4n + 4
Check that the rule makes sense with the picture. A good formula can be explained from the drawing.
The whole picture is a square with side n + 2 tiles, because there is one border tile at each end of the middle row. So the total is (n + 2)² = n² + 4n + 4.

Check. Test the rule on patterns you can see. n = 1: 1² + 4 × 1 + 4 = 1 + 4 × 1 + 4 = 1 + 4 + 4 = 9 ✓ n = 3: 3² + 4 × 3 + 4 = 9 + 4 × 3 + 4 = 9 + 12 + 4 = 25 ✓

Watch out. Do not just keep adding on a drawing for a large pattern number. Find the nth term and substitute.

(b)How many tiles are there in total in pattern 12?

Method. Count the tiles (blue and orange together), find the rule for the sequence of counts, and use the picture to see why it works.

Count the tiles (blue and orange together) in each pattern. Count carefully, one pattern at a time. This turns the pictures into a sequence of numbers.
Pattern number n 1 2 3
Number of tiles (blue and orange together) 9 16 25
Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3
Term 9 16 25
First differences: 7, 9
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: 2
They are all 2, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = 2 ÷ 2 = 1
The nth term starts with n².
Subtract n² from each term. This removes the quadratic part. Work out n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3
n² 1 4 9
n² 1 4 9
Term 9 16 25
Term − n² 8 12 16
Find the nth term of what is left. The leftover numbers form an arithmetic sequence. Find its common difference (that is the number in front of n), then the number to add, exactly as for a linear sequence.
Leftover sequence: 8, 12, 16
Common difference = 12 − 8 = 4, so it starts with 4n.
Adjustment: 8 − 4 = 4
Leftover nth term = 4n + 4
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = n² + (4n + 4) = n² + 4n + 4
Check that the rule makes sense with the picture. A good formula can be explained from the drawing.
The whole picture is a square with side n + 2 tiles, because there is one border tile at each end of the middle row. So the total is (n + 2)² = n² + 4n + 4.
Replace n by 12. Pattern 12 is the one in position 12.
Pattern 12: 12² + 4 × 12 + 4 = 144 + 4 × 12 + 4 = 144 + 48 + 4 = 196

Check. Test the rule on patterns you can see. n = 1: 1² + 4 × 1 + 4 = 1 + 4 × 1 + 4 = 1 + 4 + 4 = 9 ✓ n = 3: 3² + 4 × 3 + 4 = 9 + 4 × 3 + 4 = 9 + 12 + 4 = 25 ✓

Watch out. Do not just keep adding on a drawing for a large pattern number. Find the nth term and substitute.

Done

FFor each quadratic sequence, find the nth term and then the term asked for.

(a)4, 10, 20, 34, 52, … Find the 10th term.

Method. Find the second differences, half of them gives the number in front of n²; subtract it and the leftover is linear. Then substitute.

Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4 5
Term 4 10 20 34 52
First differences: 6, 10, 14, 18
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: 4, 4, 4
They are all 4, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = 4 ÷ 2 = 2
The nth term starts with 2n².
Subtract 2n² from each term. This removes the quadratic part. Work out 2n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4 5
n² 1 4 9 16 25
2n² 2 8 18 32 50
Term 4 10 20 34 52
Term − 2n² 2 2 2 2 2
Look at what is left. The numbers left over are all the same, so there is no n term, just a number.
Every remainder is 2, so b = 0 and c = 2.
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = 2n² + 2
Replace n by 10 to find the 10th term. The 10th term is the one in position 10.
10th term = 2 × 10² + 2 = 2 × 100 + 2 = 200 + 2 = 202

Check. Test the formula on given terms. n = 2: 2 × 2² + 2 = 2 × 4 + 2 = 8 + 2 = 10 ✓ n = 5: 2 × 5² + 2 = 2 × 25 + 2 = 50 + 2 = 52 ✓

Watch out. a is half of the second difference.

(b)2, 8, 18, 32, 50, … Find the 12th term.

Method. Find the second differences, half of them gives the number in front of n²; subtract it and the leftover is linear. Then substitute.

Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4 5
Term 2 8 18 32 50
First differences: 6, 10, 14, 18
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: 4, 4, 4
They are all 4, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = 4 ÷ 2 = 2
The nth term starts with 2n².
Subtract 2n² from each term. This removes the quadratic part. Work out 2n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4 5
n² 1 4 9 16 25
2n² 2 8 18 32 50
Term 2 8 18 32 50
Term − 2n² 0 0 0 0 0
Look at what is left. The numbers left over are all the same, so there is no n term, just a number.
Every remainder is 0, so b = 0 and c = 0.
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = 2n²
Replace n by 12 to find the 12th term. The 12th term is the one in position 12.
12th term = 2 × 12² = 2 × 144 = 288

Check. Test the formula on given terms. n = 2: 2 × 2² = 2 × 4 = 8 ✓ n = 5: 2 × 5² = 2 × 25 = 50 ✓

Watch out. a is half of the second difference.

Method. Find the second differences, half of them gives the number in front of n²; subtract it and the leftover is linear. Then substitute.

Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4 5
Term 0 3 8 15 24
First differences: 3, 5, 7, 9
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: 2, 2, 2
They are all 2, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = 2 ÷ 2 = 1
The nth term starts with n².
Subtract n² from each term. This removes the quadratic part. Work out n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4 5
n² 1 4 9 16 25
n² 1 4 9 16 25
Term 0 3 8 15 24
Term − n² −1 −1 −1 −1 −1
Look at what is left. The numbers left over are all the same, so there is no n term, just a number.
Every remainder is −1, so b = 0 and c = −1.
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = n² − 1
Replace n by 20 to find the 20th term. The 20th term is the one in position 20.
20th term = 20² − 1 = 400 − 1 = 399

Check. Test the formula on given terms. n = 2: 2² − 1 = 4 − 1 = 3 ✓ n = 5: 5² − 1 = 25 − 1 = 24 ✓

Watch out. a is half of the second difference.

Done

FIs each number a term of the sequence with the given nth term? If so, say which term.

(a)n² + 3: is 52 a term?

(b)n² + 3: is 50 a term?

(c)n² + 2n: is 80 a term?

(d)2n² + 1: is 99 a term?

(e)n² + 2n: is 79 a term?

(f)3n² − 2: is 73 a term?

Done

RSam says the nth term of 2, 6, 12, 20, … is 4n − 2. Show that Sam is wrong and find the correct nth term.

Done

RShow that the nth term of the sequence 2, 5, 10, 17, 26, … is n² + 1.

Done

RThe nth term of a sequence is n² + n. Show that every term of the sequence is an even number.

Done

RShow that the difference between the (n + 1)th square number and the nth square number is 2n + 1.

Done

PSThe nth term of a sequence is n² − 4n + 7. Work out the first five terms. Which is the smallest term?

Done

RThe first four terms of a quadratic sequence are 5, 12, 23, 38.

(a)Find the nth term.

Method. Find the second differences. Half of them gives the number in front of n². Subtract that part from the terms, and the leftover is linear.

Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
Position n 1 2 3 4
Term 5 12 23 38
First differences: 7, 11, 15
They are not all the same, so it is not a linear sequence.
Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
Second differences: 4, 4
They are all 4, so the sequence is quadratic.
Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
a = 4 ÷ 2 = 2
The nth term starts with 2n².
Subtract 2n² from each term. This removes the quadratic part. Work out 2n² for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
Position n 1 2 3 4
n² 1 4 9 16
2n² 2 8 18 32
Term 5 12 23 38
Term − 2n² 3 4 5 6
Find the nth term of what is left. The leftover numbers form an arithmetic sequence. Find its common difference (that is the number in front of n), then the number to add, exactly as for a linear sequence.
Leftover sequence: 3, 4, 5, 6
Common difference = 4 − 3 = 1, so it starts with n.
Adjustment: 3 − 1 = 2
Leftover nth term = n + 2
Put the two parts together. The nth term is the quadratic part plus the linear part.
nth term = 2n² + (n + 2) = 2n² + n + 2

Check. Test the formula. n = 2: 2 × 2² + 2 + 2 = 2 × 4 + 2 + 2 = 8 + 2 + 2 = 12 ✓ n = 4: 2 × 4² + 4 + 2 = 2 × 16 + 4 + 2 = 32 + 4 + 2 = 38 ✓

Watch out. a is half of the second difference, not the second difference itself. And remember to subtract an² (using each position) before finding the linear part.

(b)Is 500 a term of this sequence? Show how you decide.

Done

Pattern number n	1	2	3
Number of tiles (blue and orange together)	9	16	25