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Cambridge International AS & A Level Mathematics 9709 · Pure Mathematics 1

Series

The binomial expansion of $(a + b)^{n}$ for a positive integer $n$ , with the notations $(\binom{n}{r})$ and $n!$ ; arithmetic and geometric progressions, their $n$ th terms and sums; and the condition for a geometric progression to converge, with its sum to infinity.

1.6Paper 1: Pure Mathematics 1Free lesson

What the syllabus says

1.6 (Paper 1)

use the expansion of $(a + b)^{n}$ , where $n$ is a positive integer
recognise arithmetic and geometric progressions
use the formulae for the $n$ th term and for the sum of the first $n$ terms to solve problems involving arithmetic or geometric progressions
use the condition for the convergence of a geometric progression, and the formula for the sum to infinity of a convergent geometric progression.

Notes and examples

Including the notations $(\binom{n}{r})$ and $n!$ .
Knowledge of the greatest term and properties of the coefficients are not required.
Including knowledge that numbers $a$ , $b$ , $c$ are ‘in arithmetic progression’ if $2 b = a + c$ (or equivalent) and are ‘in geometric progression’ if $b^{2} = a c$ (or equivalent).
Questions may involve more than one progression.

Quoted from the Cambridge International AS and A Level Mathematics 9709 syllabus (2026-2027 and 2028-2030 editions, which share this content).

Factorials and binomial coefficients Paper 1

The factorial $n!$ is the product of the whole numbers from $n$ down to 1: $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$ . By agreement $0! = 1$ .

The binomial coefficient $(\binom{n}{r})$ , read " $n$ choose $r$ ", is the number of ways of choosing $r$ objects from $n$ , and
$(\binom{n}{r}) = \frac{n!}{r! (n - r)!}$
This is in the formula list (MF19). In practice you rarely need the full factorials, because most of the factors cancel:
$(\binom{7}{3}) = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$
There are always $r$ factors on the top, counting down from $n$ , and $r!$ underneath. Two facts save time: $(\binom{n}{0}) = (\binom{n}{n}) = 1$ , and $(\binom{n}{r}) = (\binom{n}{n - r})$ , because choosing $r$ objects to take is the same as choosing $n - r$ to leave. So $(\binom{10}{8}) = (\binom{10}{2}) = 45$ . Your calculator has an $^{n} C_{r}$ key, but the cancelling method is quicker and safer for small numbers.

The coefficients also appear as the rows of Pascal's triangle, in which each number is the sum of the two above it.

Pascal's triangle: the coefficients of $(a + b)^{n}$
$n$	Coefficients
1	1 1
2	1 2 1
3	1 3 3 1
4	1 4 6 4 1
5	1 5 10 10 5 1
6	1 6 15 20 15 6 1

Row $n$ gives the coefficients of $(a + b)^{n}$ . For example, row 4 says $(a + b)^{4} = a^{4} + 4 a^{3} b + 6 a^{2} b^{2} + 4 a b^{3} + b^{4}$ .

The binomial expansion Paper 1

For a positive integer $n$ ,
$(a + b)^{n} = a^{n} + (\binom{n}{1}) a^{n - 1} b + (\binom{n}{2}) a^{n - 2} b^{2} + \dots + (\binom{n}{r}) a^{n - r} b^{r} + \dots + b^{n}$
Why it works. Multiplying out $(a + b) (a + b) \dots (a + b)$ with $n$ brackets, every term is made by picking $a$ or $b$ from each bracket. A term $a^{n - r} b^{r}$ comes from picking $b$ in exactly $r$ of the $n$ brackets, and there are $(\binom{n}{r})$ ways of choosing those brackets. So $(\binom{n}{r})$ copies of $a^{n - r} b^{r}$ appear.

Three habits make the expansion reliable.

The powers of $a$ go down from $n$ to 0 while the powers of $b$ go up from 0 to $n$ ; in each term they add up to $n$ .
Put each part of the bracket in its own brackets before raising it to a power, including its sign and coefficient. In $(2 - 3 x)^{4}$ , $b$ is $(- 3 x)$ , so $b^{2} = 9 x^{2}$ and $b^{3} = - 27 x^{3}$ . Forgetting to raise the 3, or losing the minus sign, are the commonest errors.
The general term $(\binom{n}{r}) a^{n - r} b^{r}$ finds one coefficient without writing out the whole expansion. Write its power of $x$ in terms of $r$ and choose $r$ to give the power you want.

Questions say "in ascending powers of $x$ " (constant first) or "descending powers". "The first three terms" means the terms in $x^{0}$ , $x^{1}$ and $x^{2}$ in that order.

The syllabus does not ask for the greatest term or for properties of the coefficients.

Worked example 1 Paper 1

Expand $(2 - 3 x)^{4}$ fully, in ascending powers of $x$ .

Here $a = 2$ , $b = (- 3 x)$ and $n = 4$ , and the coefficients from Pascal's triangle are 1, 4, 6, 4, 1.
$(2 - 3 x)^{4} = 2^{4} + 4 (2^{3}) (- 3 x) + 6 (2^{2}) (- 3 x)^{2} + 4 (2) (- 3 x)^{3} + (- 3 x)^{4}$ .
$= 16 + 4 (8) (- 3 x) + 6 (4) (9 x^{2}) + 4 (2) (- 27 x^{3}) + 81 x^{4}$ .
$= 16 - 96 x + 216 x^{2} - 216 x^{3} + 81 x^{4}$ .
Check: putting $x = 1$ gives $16 - 96 + 216 - 216 + 81 = 1$ , and $(2 - 3)^{4} = 1$ .

Answer: $16 - 96 x + 216 x^{2} - 216 x^{3} + 81 x^{4}$

Worked example 2 Paper 1

Find the term independent of $x$ in the expansion of $(x^{2} + \frac{2}{x})^{6}$ .

The general term is $(\binom{6}{r}) (x^{2})^{6 - r} (\frac{2}{x})^{r} = (\binom{6}{r}) 2^{r} x^{12 - 2 r} x^{- r} = (\binom{6}{r}) 2^{r} x^{12 - 3 r}$ .
Independent of $x$ means the power of $x$ is 0: $12 - 3 r = 0$ , so $r = 4$ .
The term is $(\binom{6}{4}) 2^{4} = 15 \times 16 = 240$ .

Answer: $240$

Check yourself Paper 1

Find the coefficient of $x^{3}$ in the expansion of $(3 + 2 x)^{5}$ .

Answer it in the portal, where your answer is checked as you type →

Arithmetic progressions Paper 1

A sequence is an arithmetic progression (AP) when each term is the previous one plus a fixed common difference $d$ : for example 5, 8, 11, 14, ... with $d = 3$ . To recognise one, check that the difference between consecutive terms is constant.

With first term $a$ , the $n$ th term adds $d$ a total of $n - 1$ times:
$u_{n} = a + (n - 1) d$
The sum of the first $n$ terms is
$S_{n} = \frac{1}{2} n (a + l) = \frac{1}{2} n {2 a + (n - 1) d}$
where $l$ is the last term. Why: write the sum forwards and backwards and add. Each pair, first with last, second with second last, and so on, adds to $a + l$ , and there are $n$ pairs, so $2 S_{n} = n (a + l)$ . Both formulas are in MF19; the first is quicker when you know the last term.

Three numbers in AP. If $a$ , $b$ , $c$ are consecutive terms of an AP, then $b - a = c - b$ , which rearranges to
$2 b = a + c$
so the middle term is the mean of its neighbours.

Most AP problems give two facts and ask for $a$ and $d$ : each fact becomes an equation, and you solve them simultaneously. "Find the least $n$ for which $S_{n}$ exceeds..." leads to a quadratic inequality in $n$ ; solve the equation, then remember that $n$ must be a whole number and check the values either side.

Worked example 3 Paper 1

The fifth term of an arithmetic progression is 23 and the sum of the first 10 terms is 245.
(a) Find the first term and the common difference.
(b) Find the least number of terms for which the sum exceeds 1000.

(a) $u_{5} = a + 4 d = 23$ .
$S_{10} = \frac{1}{2} (10) (2 a + 9 d) = 245$ , so $2 a + 9 d = 49$ .
Doubling the first equation gives $2 a + 8 d = 46$ . Subtract: $d = 3$ , and then $a = 23 - 12 = 11$ .
(b) $S_{n} = \frac{1}{2} n {22 + 3 (n - 1)} = \frac{1}{2} n (3 n + 19)$ .
$\frac{1}{2} n (3 n + 19) > 1000$ gives $3 n^{2} + 19 n - 2000 > 0$ .
The positive root of $3 n^{2} + 19 n - 2000 = 0$ is $n = \frac{- 19 + \sqrt{24 361}}{6} = 22.85$ .
So $n = 23$ . Check: $S_{22} = 11 \times 85 = 935$ and $S_{23} = \frac{23}{2} \times 88 = 1012$ .

Answer: (a) $a = 11$ , $d = 3$ (b) 23 terms

Geometric progressions Paper 1

A sequence is a geometric progression (GP) when each term is the previous one multiplied by a fixed common ratio $r$ : for example 3, 6, 12, 24, ... with $r = 2$ , or 40, $- 20$ , 10, $- 5$ , ... with $r = - \frac{1}{2}$ . To recognise one, check that the ratio $\frac{u_{n + 1}}{u_{n}}$ is constant.

With first term $a$ :
$u_{n} = a r^{n - 1} S_{n} = \frac{a (1 - r^{n})}{1 - r} (r \neq 1)$
Why the sum formula works: $S_{n} = a + a r + \dots + a r^{n - 1}$ and $r S_{n} = a r + a r^{2} + \dots + a r^{n}$ . Subtracting, almost every term cancels: $S_{n} - r S_{n} = a - a r^{n}$ , so $S_{n} (1 - r) = a (1 - r^{n})$ . When $r > 1$ the equivalent form $\frac{a (r^{n} - 1)}{r - 1}$ avoids negative numbers.

Three numbers in GP. If $a$ , $b$ , $c$ are consecutive terms of a GP then $\frac{b}{a} = \frac{c}{b}$ , so
$b^{2} = a c$

Growth and decay at a fixed percentage are GPs: an amount rising by 6% a year is multiplied by $r = 1.06$ each year. Be careful which term is which year: if the first year is term 1, the 11th year is $a r^{10}$ . Questions on "the first time a term exceeds a value" lead to an inequality such as ${1.06}^{n - 1} > 2.5$ , which you solve with logarithms ( $n - 1 > \frac{\ln 2.5}{\ln 1.06}$ ) or by trying values on the calculator.

Check yourself Paper 1

Find the sum of the first 10 terms of the geometric progression 3, 6, 12, 24, ....

Answer it in the portal, where your answer is checked as you type →

Convergence and the sum to infinity Paper 1

Add up more and more terms of $8 + 4 + 2 + 1 + \frac{1}{2} + \dots$ and the total creeps towards 16 without passing it. The sum to infinity of this GP is 16.

From $S_{n} = \frac{a (1 - r^{n})}{1 - r}$ : when $- 1 < r < 1$ , $r^{n}$ gets closer and closer to 0 as $n$ increases, so $S_{n}$ gets closer and closer to $\frac{a}{1 - r}$ . The progression is said to converge, and
$S_{\infty} = \frac{a}{1 - r} provided | r | < 1$
If $| r | ⩾ 1$ the terms do not shrink, and the sum has no limit: the progression does not converge. So the condition for convergence is $| r | < 1$ , that is $- 1 < r < 1$ . Check it before using the formula, and say that you have.

When the ratio contains a variable, the condition becomes an inequality to solve. For the GP $1 + 2 x + 4 x^{2} + \dots$ the ratio is $2 x$ , so it converges when $- 1 < 2 x < 1$ , that is $- \frac{1}{2} < x < \frac{1}{2}$ .

The difference $S_{\infty} - S_{n} = \frac{a r^{n}}{1 - r}$ is what the sum still has to gain, which gives a neat way to answer "how many terms are needed to get within 0.01 of the sum to infinity?".

Worked example 4 Paper 1

A geometric progression has first term 12 and sum to infinity 48.
(a) Find the common ratio.
(b) Find the sum of the first 5 terms.

(a) $\frac{12}{1 - r} = 48$ , so $1 - r = \frac{1}{4}$ and $r = \frac{3}{4}$ . Since $| r | < 1$ , the sum to infinity does exist.
(b) $S_{5} = \frac{12 (1 - (\frac{3}{4})^{5})}{1 - \frac{3}{4}} = 48 (1 - \frac{243}{1024}) = 48 \times \frac{781}{1024} = \frac{2343}{64}$ .
As a decimal this is 36.6 (3 s.f.), comfortably below the sum to infinity, as it must be.

Answer: (a) $r = \frac{3}{4}$ (b) $\frac{2343}{64}$

Check yourself Paper 1

Find the sum to infinity of the geometric progression $20, - 5, 1.25, \dots$ .

Answer it in the portal, where your answer is checked as you type →

Check yourself Paper 1

Find the set of values of $x$ for which the geometric progression with first term 5 and common ratio $\frac{2 x}{3}$ converges.

Answer it in the portal, where your answer is checked as you type →

Problems with more than one progression Paper 1

Exam questions often link an AP and a GP: for example, three terms of an AP form a GP. Write each term of the first progression in terms of its unknowns, then use the property of the second: $b^{2} = a c$ for a GP, or $2 b = a + c$ for an AP. The result is an equation you can solve. Watch for a solution such as $d = 0$ , which the question usually rules out (an AP with $d = 0$ is constant, and then every ratio is 1).

Worked example 5 Paper 1

An arithmetic progression has first term $a$ and common difference $d$ , where $d \neq 0$ . The first, third and ninth terms of the arithmetic progression are the first three terms of a geometric progression.
(a) Show that $d = a$ .
(b) Find the common ratio of the geometric progression.

(a) The terms are $a$ , $a + 2 d$ and $a + 8 d$ . In a GP, (middle term) $^{2}$ = product of its neighbours:
$(a + 2 d)^{2} = a (a + 8 d)$ , so $a^{2} + 4 a d + 4 d^{2} = a^{2} + 8 a d$ .
So $4 d^{2} = 4 a d$ , giving $4 d (d - a) = 0$ . Since $d \neq 0$ , $d = a$ .
(b) With $d = a$ the terms are $a$ , $3 a$ and $9 a$ , so the common ratio is 3.

Answer: (a) shown (b) $r = 3$

Key points

$(\binom{n}{r}) = \frac{n!}{r! (n - r)!}$ counts the ways of choosing $r$ from $n$ ; it is the coefficient of $a^{n - r} b^{r}$ in $(a + b)^{n}$ .
Bracket each part of the binomial, with its sign and coefficient, before raising it to a power.
Use the general term $(\binom{n}{r}) a^{n - r} b^{r}$ to find a single coefficient or the term independent of $x$ .
AP: $u_{n} = a + (n - 1) d$ , $S_{n} = \frac{1}{2} n {2 a + (n - 1) d}$ ; three terms in AP satisfy $2 b = a + c$ .
GP: $u_{n} = a r^{n - 1}$ , $S_{n} = \frac{a (1 - r^{n})}{1 - r}$ ; three terms in GP satisfy $b^{2} = a c$ .
A GP converges only when $| r | < 1$ , and then $S_{\infty} = \frac{a}{1 - r}$ .
Questions asking for the least $n$ need a whole-number answer: solve, round up, and check the neighbours.

Where marks are lost

Writing $2 x^{3}$ for $(2 x)^{3}$ , or dropping the negative sign of $b$ in $(a - b x)^{n}$ .
Giving the term ( $720 x^{3}$ ) when the coefficient (720) is asked for, or the reverse.
Using $n$ instead of $n - 1$ in $u_{n}$ : the 10th term of a GP is $a r^{9}$ .
Using $S_{\infty}$ without checking that $| r | < 1$ .
Rounding $n = 22.85$ down to 22 when the sum must exceed a value.
Not rejecting a solution such as $d = 0$ or a negative ratio that the question excludes.
Miscounting years in a growth problem: the 2030 value after starting in 2020 is $a r^{10}$ , the 11th term.

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An original lesson written to the Cambridge International AS & A Level Mathematics 9709 syllabus (2026 to 2027 and 2028 to 2030). The worked examples and questions are our own, not Cambridge questions. Back to every AS & A Level Mathematics 9709 lesson.