OxfordAQA International AS and A-level Mathematics: Sequences, series and binomial expansion (9660)

This guide covers section P1.5 (Sequences and series) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. The specification states this content is part of International AS Unit P1 (Pure Maths), so it also counts towards the International A-level.

Use the revision notes for quick recall and the practice set to test yourself. The wider unit is covered in the Unit P1 study guide, and see the OxfordAQA A-level Mathematics course hub. The printable checklist lets you tick off each outcome. To find your weak spots first, try the free 10-minute diagnostics.

What this unit covers

Spec content (P1.5)	What you must be able to do
Sequences, including those given by a formula for the nth term	Generate terms, find a term, find n for a given value
Sequences generated by x(n+1) = f(x(n))	Generate terms; find a limit L by solving L = f(L)
Arithmetic series	Use the nth term and the sum; sum of the first n natural numbers; Σ notation
Sum of a finite geometric series	Use the nth term and the sum of n terms
Sum to infinity of a convergent geometric series	State and use
Binomial expansion of (1 + x)ⁿ, n a positive integer	Use n! and nCr; use Pascal’s triangle or the formula; (a + b)ⁿ accepted

Where a result below needs a formula, check which ones appear in your formulae booklet, and learn them anyway. Do not rely on being given them.

1. Sequences given by a formula for the nth term

A sequence is an ordered list of terms. If the nth term is given by a formula in n, substitute n = 1, 2, 3, … to get the terms. The specification uses the notation x(n) for a general term, and this guide writes x(n) the same way.

Worked example 1

A sequence has nth term x(n) = n² − 3n + 5.

(a) Write down the first four terms. (b) Find which term equals 59.

(a) n=1: 1 - 3 + 5 = 3
    n=2: 4 - 6 + 5 = 3
    n=3: 9 - 9 + 5 = 5
    n=4: 16 - 12 + 5 = 9
    Terms: 3, 3, 5, 9
(b) n² - 3n + 5 = 59
    n² - 3n - 54 = 0
    (n - 9)(n + 6) = 0
    n = 9 or n = -6

n must be a positive integer, so reject n = −6. The 9th term is 59. Check: 81 − 27 + 5 = 59.

2. Recurrence relations and limits

A recurrence relation gives each term from the one before: x(n+1) = f(x(n)). You also need a starting value, usually x(1).

If the sequence settles down to a limit L as n → ∞, then x(n+1) and x(n) both tend to L. Putting x(n+1) = x(n) = L gives the equation

L = f(L)

Solve it to find the limit. This tells you what the limit would be if there is one. It does not prove the sequence converges.

Worked example 2

x(n+1) = 0.6 x(n) + 8, with x(1) = 5.

(a) Find x(2), x(3), x(4). (b) Find the limit.

(a) x(2) = 0.6(5) + 8 = 11
    x(3) = 0.6(11) + 8 = 14.6
    x(4) = 0.6(14.6) + 8 = 16.76
(b) L = 0.6L + 8
    0.4L = 8
    L = 20

The terms are increasing and approach 20.

Worked example 3 (more than one root)

x(n+1) = √(x(n) + 12), with x(1) = 1.

L = √(L + 12)
L² = L + 12
L² - L - 12 = 0
(L - 4)(L + 3) = 0
L = 4 or L = -3

Squaring can introduce a false root. Every term is a positive square root, so the limit cannot be negative. Reject L = −3, so L = 4. The first terms are 1, 3.606, 3.950, 3.994 (3 d.p.), which agree with 4.

Some recurrences never settle. For example, x(n+1) = 1/(1 − x(n)) with x(1) = 3 gives 3, −1/2, 2/3, 3, … The sequence repeats every three terms. It is periodic, so it has no limit. To find a late term, use the period: x(100) is the same as x(1), because 100 leaves remainder 1 when divided by 3.

3. Arithmetic series

In an arithmetic sequence you add the same common difference d each time. With first term a:

nth term:   x(n) = a + (n - 1)d
sum:        S(n) = n/2 × (2a + (n - 1)d)  =  n/2 × (a + l)

Here l is the last term. Use the second form when you know the last term.

Sum of the first n natural numbers

Take a = 1, d = 1:

1 + 2 + 3 + ... + n = n(n + 1)/2

For example, 1 + 2 + … + 100 = 100 × 101/2 = 5050.

Σ notation

Σ means “add up”. The expression Σ from r = 1 to 24 of (5r − 2) means: put r = 1, 2, …, 24 into 5r − 2 and add. It is an arithmetic series with first term 3, common difference 5 and 24 terms.

S = 24/2 × (3 + 118) = 12 × 121 = 1452

Check using the natural-number result: 5 × (24 × 25/2) − 2 × 24 = 5 × 300 − 48 = 1452.

Worked example 4

The 5th term of an arithmetic sequence is 23 and the 12th term is 58.

(a) Find a and d. (b) Find the sum of the first 20 terms. (c) Find the smallest n for which the sum of the first n terms exceeds 500.

(a) a + 4d = 23
    a + 11d = 58
    Subtract: 7d = 35, so d = 5 and a = 3.
(b) S(20) = 20/2 × (2(3) + 19(5)) = 10 × 101 = 1010
(c) S(n) = n/2 × (6 + 5(n - 1)) = n(5n + 1)/2
    n(5n + 1)/2 > 500
    5n² + n - 1000 > 0

The positive root of 5n² + n − 1000 = 0 is n = (−1 + √20001)/10 ≈ 14.04. So n must be at least 15. Check both sides of the boundary: S(14) = 497, which is not above 500, and S(15) = 570. The answer is n = 15.

4. Geometric series

In a geometric sequence you multiply by the same common ratio r each time. With first term a:

nth term:   x(n) = a rⁿ⁻¹
sum:        S(n) = a(1 - rⁿ)/(1 - r)  =  a(rⁿ - 1)/(r - 1),   r ≠ 1

Use whichever form has the positive denominator. The sum comes from subtracting rS from S, so almost all terms cancel.

Worked example 5

A geometric series has first term 4 and common ratio 1.5.

(a) Find the sum of the first 8 terms. (b) Find the smallest n for which S(n) exceeds 10 000.

(a) S(8) = 4(1.5⁸ - 1)/(1.5 - 1) = 8(1.5⁸ - 1)
    1.5⁸ = 6561/256
    S(8) = 8 × 6305/256 = 6305/32 = 197.03125
(b) 8(1.5ⁿ - 1) > 10 000
    1.5ⁿ > 1251
    n > ln 1251 / ln 1.5 = 17.6...

Because n must be a whole number, n = 18. Check: S(17) = 7874.09… and S(18) = 11 815.1…, so 17 terms is not enough.

5. Sum to infinity

If |r| < 1, that is −1 < r < 1, the terms shrink towards zero and the sum of n terms approaches a limit as n → ∞. In S(n), the rⁿ term vanishes, leaving

S∞ = a/(1 - r),     |r| < 1

If |r| ≥ 1 the series does not converge, so there is no sum to infinity.

Worked example 6

A geometric series has all terms positive. Its 2nd term is 12 and its 4th term is 3.

ar = 12 and ar³ = 3
Divide: r² = 3/12 = 1/4
r = 1/2 or r = -1/2

All terms are positive, so reject r = −1/2. Then r = 1/2 and a = 24. The series converges because |r| < 1, and S∞ = 24/(1 − 1/2) = 48.

Worked example 7 (range of x)

Find the values of x for which 1 + (2x − 1) + (2x − 1)² + … converges, and its sum when it does.

The ratio is r = 2x − 1. Convergence needs |2x − 1| < 1, so −1 < 2x − 1 < 1, which gives 0 < x < 1. The sum is 1/(1 − (2x − 1)) = 1/(2 − 2x).

Recurring decimals

0.272727… = 0.27 + 0.0027 + … is geometric with a = 0.27 and r = 0.01. So S∞ = 0.27/0.99 = 27/99 = 3/11.

6. The binomial expansion

Factorials and nCr

n! = n × (n − 1) × … × 2 × 1, with 0! = 1. The number of ways to choose r items from n is written nCr (the specification also shows it as a bracket with n over r):

nCr = n! / (r!(n - r)!)

For example, 7C3 = 7!/(3! 4!) = 35, and nC0 = nCn = 1. Also nCr = nC(n−r), so the coefficients are symmetric.

Pascal’s triangle

Each row begins and ends with 1, and every other entry is the sum of the two above it. Row n gives the coefficients of (1 + x)ⁿ.

n=0:           1
n=1:          1 1
n=2:         1 2 1
n=3:        1 3 3 1
n=4:       1 4 6 4 1
n=5:     1 5 10 10 5 1
n=6:   1 6 15 20 15 6 1

The expansion

For positive integer n:

(1 + x)ⁿ = 1 + nx + [n(n-1)/2!] x² + [n(n-1)(n-2)/3!] x³ + ... + xⁿ

For (a + b)ⁿ, the term in aⁿ⁻ʳbʳ has coefficient nCr, and the powers of a and b in each term add up to n. The specification says Pascal’s triangle or formulae will be accepted, so use whichever is faster and safer for you.

Worked example 8

Expand (2 − 3x)⁵.

The coefficients are 1, 5, 10, 10, 5, 1. Take a = 2 and b = −3x.

32                    =  32
5 × 2⁴ × (-3x)        =  5 × 16 × (-3x)    = -240x
10 × 2³ × (-3x)²      =  10 × 8 × 9x²      =  720x²
10 × 2² × (-3x)³      =  10 × 4 × (-27x³)  = -1080x³
5 × 2 × (-3x)⁴        =  5 × 2 × 81x⁴      =  810x⁴
(-3x)⁵                =                      -243x⁵

So (2 − 3x)⁵ = 32 − 240x + 720x² − 1080x³ + 810x⁴ − 243x⁵. Put x = 1 as a check: (−1)⁵ = −1 and the coefficients sum to 32 − 240 + 720 − 1080 + 810 − 243 = −1.

Worked example 9 (a single coefficient)

Find the coefficient of x³ in (1 + 2x)⁷. The x³ term is 7C3 × (2x)³ = 35 × 8x³ = 280x³. The coefficient is 280.

Worked example 10 (a product)

Find the coefficient of x² in (3 + x)(1 − 2x)⁶. First, (1 − 2x)⁶ = 1 − 12x + 60x² − 160x³ + … For x², take 3 × 60x² from the first bracket’s 3 and x × (−12x) from its x:

3 × 60 + (-12) = 180 - 12 = 168

The coefficient is 168.

Worked example 11 (approximation)

Use the first three terms of (1 + x)⁶ with x = 0.02 to estimate 1.02⁶.

1 + 6(0.02) + 15(0.02)² = 1 + 0.12 + 0.006 = 1.126

The exact value is 1.126162… so the estimate is a little low, because the omitted terms are positive.

Common errors

• Using n for the number of terms and also for the power of r. In x(n) = arⁿ⁻¹ the power is n − 1, not n.
• Forgetting to reject a root. A limit must suit the recurrence, as in example 3, and n must be a positive integer.
• Writing L = f(L) and stopping. Check the answer is sensible, and remember it only applies if the sequence converges.
• Quoting S∞ = a/(1 − r) without checking |r| < 1.
• In inequality questions, rounding n down. If S(14) falls short, the answer is 15, not 14.
• Dropping brackets on a negative term, for example (−3x)² is 9x², not −9x².
• Leaving out the coefficient when b has a number in front of x: (2x)³ is 8x³, so the 8 must multiply the nCr value.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, published by OxfordAQA: section P1.5, Sequences and series, under International AS Unit P1 (Pure Maths).