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

Quick recall for section P1.5 of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. The specification places it in International AS Unit P1 (Pure Maths). For full explanations, read the study guide, then test yourself with the practice set.

Related pages: the Unit P1 revision notes, the course hub, the printable checklist and the free 10-minute diagnostics.

Formulae to learn

Check your formulae booklet to see which of these it gives, but know them all.

Topic	Result
Arithmetic nth term	x(n) = a + (n − 1)d
Arithmetic sum	S(n) = n/2 × (2a + (n − 1)d) = n/2 × (a + l)
Sum of 1 to n	1 + 2 + … + n = n(n + 1)/2
Geometric nth term	x(n) = a rⁿ⁻¹
Geometric sum	S(n) = a(1 − rⁿ)/(1 − r), r ≠ 1
Sum to infinity	S∞ = a/(1 − r), only when −1 < r < 1
Limit of x(n+1) = f(x(n))	Solve L = f(L)
Factorial	n! = n(n − 1)…2 × 1, and 0! = 1
Choose	nCr = n!/(r!(n − r)!), with nC0 = nCn = 1
Binomial	(a + b)ⁿ = Σ nCr aⁿ⁻ʳ bʳ, r from 0 to n

Must-know distinctions

	Arithmetic	Geometric
Rule	Add the common difference d	Multiply by the common ratio r
Test	Differences are equal	Ratios are equal
nth term	a + (n − 1)d	a rⁿ⁻¹
Sum to infinity	Never exists (unless every term is 0)	Exists only if

A sequence is the list of terms. A series is the sum of the terms. Σ notation describes a series.

Which result do I use?

The question gives you	Reach for
A formula in n	Substitution, or solve formula = value
x(n+1) in terms of x(n)	Repeated substitution; L = f(L) for a limit
A constant difference	Arithmetic formulae
A constant ratio	Geometric formulae
“Sum to infinity” or “converges”
A bracket raised to a small power	Pascal’s triangle
A bracket raised to a larger power, or one coefficient	nCr

Sequences from a formula

Substitute n = 1, 2, 3, … To find which term equals a value, set the formula equal to it and solve. Keep only positive integer solutions of n.

Worked reminder: x(n) = n(n + 2). The first terms are 3, 8, 15. To find which term is 120, solve n² + 2n − 120 = 0, so (n + 12)(n − 10) = 0. Reject n = −12, so it is the 10th term. If a solution is not a positive integer, the value is not a term of the sequence.

Recurrence relations: method

1. Write x(n+1) = f(x(n)) and the start value x(1).
2. Find terms by repeated substitution. Keep full accuracy until the end.
3. For a limit, replace both x(n+1) and x(n) by L.
4. Solve L = f(L).
5. Reject any root that cannot fit the sequence (a negative value of a
   square root, for example).

Worked reminder: x(n+1) = 0.75x(n) + 7. Put L = 0.75L + 7, so 0.25L = 7 and L = 28.

A recurrence that cycles (for example 3, −1/2, 2/3, 3, …) has a period. It has no limit.

Arithmetic series: method

1. Write down what you know: a, d, n, l, S(n).
2. Pick the formula that contains exactly the unknown you want.
3. For two conditions, form two equations in a and d and solve.
4. For "least n" questions, form an inequality, solve the boundary,
   then test the whole numbers either side.

Worked reminder: a = 10, d = −3. Find the sum of the 6th to 15th terms.

S(15) = 15/2 × (20 + 14(-3)) = 15/2 × (-22) = -165
S(5)  = 5/2 × (20 + 4(-3))  = 5/2 × 8        = 20
Sum of 6th to 15th = S(15) - S(5) = -185

The same idea works for any block of terms: subtract the first part from the whole.

Σ notation

Σ from r = 1 to N of (pr + q) is arithmetic with first term p + q and common difference p. Split sums when it helps: Σ(pr + q) = pΣr + qN, using Σr = N(N + 1)/2.

Worked reminder: Σ from r = 1 to 10 of (3r + 2) = 3 × 55 + 2 × 10 = 185.

Note that the lower limit matters. If the sum starts at r = 6, subtract the first five terms from the total.

Geometric series: method

1. Find r by dividing a term by the one before it.
2. Use ar^(n-1) for a term and a(1 - r^n)/(1 - r) for a sum.
3. "Sum exceeds a value" needs an inequality in r^n.
   Take logs and keep the inequality the right way round
   (dividing by ln r flips it only if ln r < 0).
4. If two terms are given, divide them to get a power of r.

Worked reminder: the 3rd term is 20 and the 6th term is 160.

ar² = 20 and ar⁵ = 160
r³ = 160/20 = 8, so r = 2
a = 20/4 = 5
S(8) = 5(2⁸ - 1)/(2 - 1) = 5 × 255 = 1275

If r could be positive or negative, use any extra information (all terms positive, for instance) to choose.

Sum to infinity

S∞ = a/(1 - r)    valid only for |r| < 1, that is -1 < r < 1

Worked reminder: a = 81, r = 1/3 gives S∞ = 81/(2/3) = 121.5.

For “find the range of x for which the series converges”, write the ratio r in terms of x and solve |r| < 1 as a double inequality.

Recurring decimals: write as a geometric series. For 0.4545… use a = 0.45, r = 0.01, so S∞ = 0.45/0.99 = 5/11.

Binomial expansion: method

1. Choose Pascal's triangle (small n) or nCr (larger n).
2. Write each term as nCr × (first)^(n-r) × (second)^r.
3. Put brackets around negative or numerical terms, then simplify.
4. Check by substituting a simple value such as x = 1.

Pascal’s triangle rows:

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

Worked reminder: (x − 2)⁴ = x⁴ − 8x³ + 24x² − 32x + 16. The x³ term is 4 × x³ × (−2) = −8x³.

Worked reminder: the coefficient of x² in (1 + kx)⁵ is 90. Find k.

5C2 × k² = 90
10k² = 90
k² = 9, so k = 3 or k = -3

Both values satisfy the condition unless the question says k > 0 or similar. Read the wording for any restriction before you discard a root.

For a product such as (p + qx)(1 + sx)ⁿ, find each combination of terms that gives the power you want, then add the contributions.

For (1 + x)ⁿ, the first three terms are 1 + nx + n(n − 1)/2 × x². They give a good estimate when x is small.

Quick self-test

Try each before you look at the answers.

1. A sequence has x(n) = 4n − 7. Find x(10).
2. x(n+1) = x(n)/2 + 3. Find the limit.
3. An arithmetic series has a = 2 and d = 3. Find S(15).
4. Find 1 + 2 + 3 + … + 60.
5. A geometric series has a = 3 and r = 2. Find S(7).
6. A geometric series has a = 18 and r = −1/3. Find S∞.
7. Evaluate 9C4.
8. Find the coefficient of x² in (1 + 3x)⁵.
9. Expand (1 − x)⁴.
10. State the values of r for which a geometric series converges.
11. Evaluate Σ from r = 1 to 6 of 2^r.

Answers

1. x(10) = 40 − 7 = 33.
2. L = L/2 + 3, so L/2 = 3 and L = 6.
3. S(15) = 15/2 × (4 + 14 × 3) = 15/2 × 46 = 345.
4. 60 × 61/2 = 1830.
5. S(7) = 3(2⁷ − 1)/(2 − 1) = 3 × 127 = 381.
6. S∞ = 18/(1 + 1/3) = 18/(4/3) = 27/2, which is 13.5.
7. 9C4 = 9!/(4! 5!) = 126.
8. 5C2 × 3² = 10 × 9 = 90.
9. 1 − 4x + 6x² − 4x³ + x⁴.
10. −1 < r < 1, written |r| < 1.
11. Geometric with a = 2, r = 2, six terms: 2(2⁶ − 1)/(2 − 1) = 126.

Where marks are usually lost

• Taking n = 0 as the first term, or mixing up “the nth term” with “the sum of n terms”.
• Using a rⁿ instead of a rⁿ⁻¹ for the nth term.
• Applying S∞ = a/(1 − r) without showing |r| < 1.
• Leaving a false root from squaring in a limit question.
• Rounding n to the nearest whole number instead of the next whole number that satisfies the inequality.
• Losing a negative sign in (−2)³ or (−3x)².
• Forgetting that nCr multiplies the numerical coefficient too, as in (2x)³ = 8x³.
• Stopping a Σ question at the formula without evaluating, or ignoring a lower limit other than 1.
• Not giving an exact answer when the question asks for one.

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).