Series

Syllabus coverage

Use the expansion of $(a + b)^{n}$ , where $n$ is a positive integer, including the notations $(\binom{n}{r})$ and $n!$
Recognise arithmetic and geometric progressions
Use the formulae for the nth term and for the sum of the first $n$ terms to solve problems involving arithmetic or geometric progressions, including the facts that $a$ , $b$ , $c$ are in arithmetic progression when $2 b = a + c$ and in geometric progression when $b^{2} = a c$
Use the condition for the convergence of a geometric progression, and the formula for the sum to infinity of a convergent geometric progression; questions may involve more than one progression

Part of the AS & A Level Maths syllabus 9709.

An arithmetic progression has a constant step

d

, so the terms rise in a straight line; a geometric progression multiplies by

r

each time. When

| r | < 1

the geometric partial sums creep up towards

\frac{a}{1 - r}

but never reach it.

Factorials and binomial coefficients

The factorial of a positive integer is the product of every whole number down to 1: $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$ , and in general $n! = n (n - 1) (n - 2) \dots \times 2 \times 1$ . Two values have to be known by definition rather than by multiplication: $0! = 1$ , and $1! = 1$ . Factorials grow very fast, which is why the numbers in this topic get large very quickly: $10! = 3 628 800$ .

A binomial coefficient counts the number of ways of choosing $r$ objects from $n$ when the order does not matter, and it is written $(\binom{n}{r})$ or $^{n} C_{r}$ , both read as 'n choose r'. The formula is $(\binom{n}{r}) = \frac{n!}{r! (n - r)!}$ . So $(\binom{6}{2}) = \frac{6!}{2! 4!} = \frac{720}{2 \times 24} = 15$ , and $(\binom{10}{5}) = \frac{10!}{5! 5!} = 252$ .

Three facts save a lot of arithmetic. First, $(\binom{n}{0}) = (\binom{n}{n}) = 1$ , because there is only one way to choose nothing or everything. Second, $(\binom{n}{1}) = n$ . Third, $(\binom{n}{r}) = (\binom{n}{n - r})$ , which is why Pascal's triangle is symmetrical: the coefficient of $x^{2}$ and the coefficient of $x^{4}$ in the expansion of $(1 + x)^{6}$ are both 15. Your calculator has an nCr button, and using it is fine, but write the notation in your working because the mark scheme looks for it.

$n$	The coefficients of $(a + b)^{n}$
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

Each row of Pascal's triangle is built from the row above it: add the two numbers either side to get the number below. The row for

n = 7

starts 1, 7, 21, 35, and carries on 35, 21, 7, 1.

Expanding $(a + b)^{n}$

For a positive integer $n$ , $(a + b)^{n} = a^{n} + (\binom{n}{1}) a^{n - 1} b + (\binom{n}{2}) a^{n - 2} b^{2} + \dots + b^{n}$ . The pattern is fixed: the powers of $a$ count down from $n$ to 0, the powers of $b$ count up from 0 to $n$ , the two powers always add to $n$ , and the coefficients come from row $n$ of Pascal's triangle. There are $n + 1$ terms altogether.

So $(1 + 2 x)^{4} = 1 + 4 (2 x) + 6 (2 x)^{2} + 4 (2 x)^{3} + (2 x)^{4} = 1 + 8 x + 24 x^{2} + 32 x^{3} + 16 x^{4}$ . Notice that the whole of $2 x$ , bracket and all, is raised to the power: it is $(2 x)^{2} = 4 x^{2}$ , not $2 x^{2}$ . The general term is $(\binom{n}{r}) a^{n - r} b^{r}$ , which is the term containing $b^{r}$ .

A minus sign inside the bracket makes the signs alternate. $(2 - x)^{5} = 2^{5} + 5 (2^{4}) (- x) + 10 (2^{3}) (- x)^{2} + 10 (2^{2}) (- x)^{3} + 5 (2) (- x)^{4} + (- x)^{5}$ , that is $32 - 80 x + 80 x^{2} - 40 x^{3} + 10 x^{4} - x^{5}$ . Write the bracket round the negative term at every stage, so that the odd powers keep their minus sign.

Questions rarely ask for the whole expansion. They ask for one coefficient, and the quick way is to write down only the term you need. The coefficient of $x^{2}$ in $(1 + 3 x)^{6}$ comes from $(\binom{6}{2}) (3 x)^{2} = 15 \times 9 x^{2} = 135 x^{2}$ , so the coefficient is 135. The coefficient of $x^{5}$ in $(x + 2)^{10}$ comes from $(\binom{10}{5}) x^{5} 2^{5} = 252 \times 32 x^{5} = 8064 x^{5}$ , so the coefficient is 8064. Notice how the roles of the two parts swap between those two examples: identify which part carries the $x$ before you decide which power to take.

$r$	$(\binom{4}{r})$	Term of $(1 + 2 x)^{4}$
0	1	$1$
1	4	$8 x$
2	6	$24 x^{2}$
3	4	$32 x^{3}$
4	1	$16 x^{4}$

The coefficient of

x^{r}

is not just the binomial coefficient. Every part of the term contributes: in

(1 + 3 x)^{6}

the coefficient of

x^{2}

15 \times 9

, and forgetting the

3^{2}

is the commonest way to lose that mark.

Arithmetic progressions

A sequence is a list of numbers in order, and a series is what you get when you add the terms of a sequence. A sequence is an arithmetic progression when you add the same amount each time to get the next term; that amount is the common difference, written $d$ . The progression with first term $a$ and common difference $d$ is $a$ , $a + d$ , $a + 2 d$ , $a + 3 d$ , and the nth term is $u_{n} = a + (n - 1) d$ . The $n - 1$ is not a detail you can drop: to reach the 20th term you add $d$ nineteen times.

The sum of the first $n$ terms is $S_{n} = \frac{n}{2} {2 a + (n - 1) d}$ , which is the version to use when you know $a$ and $d$ . If you know the last term $l$ instead, $S_{n} = \frac{n}{2} (a + l)$ does the same job more quickly. Both formulae are in the formula booklet, and both come from the same idea: pairing the first term with the last, the second with the second last, and so on, each pair having the same total.

Three numbers $a$ , $b$ , $c$ are in arithmetic progression when the middle one is the average of the other two, that is when $2 b = a + c$ . So 7, 11 and 15 are in arithmetic progression because $2 \times 11 = 7 + 15$ , and the value $b$ is always the mean of $a$ and $c$ .

Most exam questions give you two terms and ask for $a$ and $d$ . Write each term with the formula and subtract the two equations: the $a$ cancels and $d$ drops out. For a progression with $u_{5} = 22$ and $u_{12} = 57$ , the equations are $a + 4 d = 22$ and $a + 11 d = 57$ ; subtracting gives $7 d = 35$ , so $d = 5$ , and then $a = 22 - 4 (5) = 2$ . The progression is 2, 7, 12, 17, ... .

Quantity	Formula	For the progression 7, 12, 17, ...
nth term	$u_{n} = a + (n - 1) d$	$u_{n} = 7 + 5 (n - 1) = 5 n + 2$
Sum of the first $n$ terms	$S_{n} = \frac{n}{2} {2 a + (n - 1) d}$	$S_{20} = \frac{20}{2} {14 + 19 (5)} = 1090$
Sum using the last term	$S_{n} = \frac{n}{2} (a + l)$	$S_{20} = \frac{20}{2} (7 + 102) = 1090$

To reach the 20th term of 7, 12, 17, ... you add 5 nineteen times, so

u_{20} = 7 + 19 (5) = 102

. Adding it twenty times gives 107 and is wrong, which is why the formula has

n - 1

in it.

Geometric progressions

A sequence is a geometric progression when you multiply by the same number each time to get the next term; that number is the common ratio, written $r$ . The progression with first term $a$ and common ratio $r$ is $a$ , $a r$ , $a r^{2}$ , $a r^{3}$ , so the nth term is $u_{n} = a r^{n - 1}$ . Again the index is $n - 1$ , not $n$ .

The sum of the first $n$ terms is $S_{n} = \frac{a (r^{n} - 1)}{r - 1}$ for $r \neq 1$ , and the equivalent form $S_{n} = \frac{a (1 - r^{n})}{1 - r}$ is easier to use when $r$ is a fraction, because then both top and bottom are positive. When $r = 1$ the formula breaks down (it would divide by zero) and every term is $a$ , so $S_{n} = n a$ .

Three numbers $a$ , $b$ , $c$ are in geometric progression when $b^{2} = a c$ , which is the same as saying that $b$ is the geometric mean of $a$ and $c$ . So 4, 6 and 9 are in geometric progression because $6^{2} = 36 = 4 \times 9$ , with common ratio $\frac{3}{2}$ .

When two terms of a geometric progression are given, divide rather than subtract. The division cancels the unknown $a$ and leaves a power of $r$ . If the second term is 6 and the fifth is 48, then $a r = 6$ and $a r^{4} = 48$ ; dividing gives $r^{3} = 8$ , so $r = 2$ and $a = \frac{6}{2} = 3$ . The progression is 3, 6, 12, 24, 48, ... . A progression with first term 2 and ratio 3 has $u_{10} = 2 \times 3^{9} = 39 366$ and $S_{10} = \frac{2 (3^{10} - 1)}{3 - 1} = 59 048$ .

Quantity	Formula	For the progression 2, 6, 18, ...
nth term	$u_{n} = a r^{n - 1}$	$u_{n} = 2 \times 3^{n - 1}$
Sum of the first $n$ terms	$S_{n} = \frac{a (r^{n} - 1)}{r - 1}$	$S_{10} = \frac{2 (3^{10} - 1)}{2} = 59 048$
Sum to infinity, when $\| r \| < 1$	$S_{\infty} = \frac{a}{1 - r}$	not applicable, since $r = 3$

The common ratio can be negative or a fraction. The progression 8,

- 4

, 2,

- 1

, ... has

r = - \frac{1}{2}

, and its terms alternate in sign while shrinking towards zero.

Convergence and the sum to infinity

A geometric progression converges when its terms get closer and closer to zero as $n$ increases, and that happens exactly when $| r | < 1$ . Then $r^{n} \to 0$ as $n \to \infty$ , so the sum of the first $n$ terms settles down to a fixed value, called the sum to infinity: $S_{\infty} = \frac{a}{1 - r}$ . The condition is on the size of $r$ , not on its sign, so $r = - 0.9$ converges just as $r = 0.9$ does.

If $| r | \geq 1$ the terms do not shrink, so the sum has no limiting value and the progression is divergent. For $r = 1$ every term is $a$ and the sum grows without limit; for $r = - 1$ the sum alternates between $a$ and 0 and never settles; for $r = 2$ the terms double and the sum runs away. In each of those cases $S_{\infty} = \frac{a}{1 - r}$ gives a number that means nothing, and quoting it is a mistake even if the arithmetic is right.

So 8 + 4 + 2 + 1 + $\dots$ has $a = 8$ and $r = \frac{1}{2}$ , and its sum to infinity is $\frac{8}{1 - \frac{1}{2}} = \frac{8}{\frac{1}{2}} = 16$ . The partial sums are 8, 12, 14, 15, 15.5, ... and they creep towards 16 without ever reaching it, which is exactly what converging means.

Recurring decimals are geometric progressions in disguise, and this is where the formula earns its keep. $0.444 \dots = 0.4 + 0.04 + 0.004 + \dots$ , a geometric progression with $a = 0.4$ and $r = 0.1$ , so its sum is $\frac{0.4}{1 - 0.1} = \frac{0.4}{0.9} = \frac{4}{9}$ . The same method converts any recurring decimal into a fraction, and it is the neatest proof that a recurring decimal is rational.

Common ratio $r$	What the terms do	Sum to infinity
$\| r \| < 1$	shrink towards zero	$S_{\infty} = \frac{a}{1 - r}$ exists
$r = 0$	$a$ , then zeros for ever	$S_{\infty} = a$
$r = 1$	every term is $a$	none: $S_{n} = n a$ grows without limit
$r = - 1$	alternate $a$ , $- a$ , $a$ , $- a$ , ...	none: the sum never settles
$\| r \| > 1$	grow in size without limit	none: the progression diverges

Before you use

S_{\infty} = \frac{a}{1 - r}

, write down that

| r | < 1

. That one line is often a mark of its own, and it is the step candidates skip when they quote the formula for a progression that diverges.

Tackling a series problem

Almost every question in this topic is one of four shapes, and each has a first move. If you are given two terms, write both with the nth term formula and then either subtract (arithmetic) or divide (geometric). If you are given a sum, use the appropriate $S_{n}$ formula and solve for what is missing. If you are asked how many terms are needed for a given sum, form an equation in $n$ , which for an arithmetic progression is a quadratic. If you are given a condition such as 'three consecutive terms are in geometric progression', use $b^{2} = a c$ .

The 'how many terms' question is where careless answers appear. $5 + 9 + 13 + \dots$ has $a = 5$ and $d = 4$ , and asking for the number of terms with sum 1325 means solving $\frac{n}{2} {10 + 4 (n - 1)} = 1325$ , which tidies to $2 n^{2} + 3 n - 1325 = 0$ and factorises as $(2 n + 53) (n - 25) = 0$ . The two roots are $n = 25$ and $n = - 26.5$ ; only the positive whole number can be the number of terms, so $n = 25$ . Always say why the other root is rejected, because that line is a mark.

A question may involve two progressions at once, and the trick is to work out what each one is being asked for before equating anything. If one progression grows by a fixed amount each year and another by a fixed percentage, the first is arithmetic and the second geometric; 'when are they equal' then means comparing $a + (n - 1) d$ with $a r^{n - 1}$ , and 'when has the first overtaken the second' means finding the smallest whole $n$ that makes the inequality true.

Finally, keep the words 'term' and 'sum' apart. The nth term is one number in the list; the sum of the first $n$ terms is everything added up to that point. A question that says 'the total after 10 years' wants $S_{10}$ , and a question that says 'the amount saved in the 10th year' wants $u_{10}$ . Reading that difference correctly is worth more marks than any algebra in this topic.

Two terms given: subtract for an arithmetic progression, divide for a geometric one
A sum given, terms asked for: write $S_{n}$ equal to the sum and solve, keeping only the sensible root
Three consecutive terms in geometric progression: use $b^{2} = a c$
Fixed amount added each step means arithmetic; fixed percentage means geometric
A sum grows by a percentage each year, so it is the terms of a geometric progression you add

'Diverges' and 'does not converge' are complete answers when a question asks whether a sum to infinity exists. If

| r | \geq 1

, say so, and do not try to compute a value.

Key terms you must be able to define

Sequence: A list of numbers in a definite order, such as 3, 6, 12, 24. The order matters, and each number in the list is a term.

Series: The sum of the terms of a sequence: 3 + 6 + 12 + 24. A question about a series wants a total.

nth term: The term in position $n$ of a sequence, written $u_{n}$ . It is one number, not a total.

Arithmetic progression: A sequence in which the difference between consecutive terms is constant, such as 7, 12, 17, 22, ... with common difference 5.

Common difference: The constant amount added to each term of an arithmetic progression to get the next one, written $d$ .

Geometric progression: A sequence in which each term is a constant multiple of the one before it, such as 2, 6, 18, 54, ... with common ratio 3.

Common ratio: The constant multiplier between consecutive terms of a geometric progression, written $r$ . It may be negative or a fraction.

Convergent: A geometric progression with $| r | < 1$ , whose terms tend to zero and whose sum therefore tends to a finite limit.

Divergent: A geometric progression with $| r | \geq 1$ , whose terms do not tend to zero, so no sum to infinity exists.

Sum to infinity: The limiting value of the sum of a convergent geometric progression, $S_{\infty} = \frac{a}{1 - r}$ with $| r | < 1$ .

Factorial: $n! = n (n - 1) (n - 2) \dots \times 2 \times 1$ , the product of all the whole numbers from 1 up to $n$ . By definition $0! = 1$ .

Binomial coefficient: $(\binom{n}{r}) = \frac{n!}{r! (n - r)!}$ , the coefficient of $a^{n - r} b^{r}$ in the expansion of $(a + b)^{n}$ .

Pascal's triangle: The triangle of binomial coefficients, each row beginning and ending with 1 and built by adding the two numbers above.

Binomial expansion: The expansion of $(a + b)^{n}$ as a sum of $n + 1$ terms, the general term being $(\binom{n}{r}) a^{n - r} b^{r}$ .

General term: The term containing $b^{r}$ in the expansion of $(a + b)^{n}$ , namely $(\binom{n}{r}) a^{n - r} b^{r}$ . Writing it down is the quick way to find a single coefficient.

Applied

At AS & A Level the marks sit in the setup: choosing the model, stating the variable and writing the equation before any working. These show the shape of a full-mark answer, then you do the algebra yourself.

A binomial coefficient worked by hand

Calculate $(\binom{7}{4})$ without a calculator. Using the formula, $(\binom{7}{4}) = \frac{7!}{4! 3!}$ . Rather than writing out all the factorials, cancel the $4!$ against the top: $\frac{7 \times 6 \times 5}{3 \times 2 \times 1} = \frac{210}{6} = 35$ . The check is symmetry: $(\binom{7}{3}) = 35$ too, because choosing 4 objects to keep is the same as choosing 3 to leave out.

Expanding a bracket with a minus sign

Expand $(2 - x)^{5}$ . The coefficients from row 5 of Pascal's triangle are 1, 5, 10, 10, 5, 1, and the powers of 2 count down while the powers of $- x$ count up: $2^{5} + 5 (2^{4}) (- x) + 10 (2^{3}) (- x)^{2} + 10 (2^{2}) (- x)^{3} + 5 (2) (- x)^{4} + (- x)^{5}$ . That is $32 - 80 x + 80 x^{2} - 40 x^{3} + 10 x^{4} - x^{5}$ . Every odd power of $x$ carries a minus sign.

An arithmetic progression from two terms

The fourth term of an arithmetic progression is 17 and the ninth term is 42. Then $a + 3 d = 17$ and $a + 8 d = 42$ . Subtracting the first equation from the second gives $5 d = 25$ , so $d = 5$ , and $a = 17 - 3 (5) = 2$ . The progression is 2, 7, 12, 17, 22, ... and its twentieth term is $2 + 19 (5) = 97$ .

A geometric progression and its sum to infinity

For the progression 8, 4, 2, 1, ..., the first term is 8 and the common ratio is $\frac{1}{2}$ . Since $| \frac{1}{2} | < 1$ , the progression converges and $S_{\infty} = \frac{8}{1 - \frac{1}{2}} = 16$ . The sum of the first 10 terms is $\frac{8 (1 - {0.5}^{10})}{1 - 0.5} = 15.984$ , which is close to 16 but not equal to it: the sum to infinity is a limit, never a term of the sequence.

A recurring decimal as a geometric progression

Write $0.777 \dots$ as a fraction. The decimal is $0.7 + 0.07 + 0.007 + \dots$ , a geometric progression with $a = 0.7$ and $r = 0.1$ . Since $| r | < 1$ , $S_{\infty} = \frac{0.7}{1 - 0.1} = \frac{0.7}{0.9} = \frac{7}{9}$ . The answer is exact, and the method works for any recurring block of digits.

Flashcards

Click a card to flip it. Mark honestly — the tracker uses this.

Q1What is

5!

5 \times 4 \times 3 \times 2 \times 1 = 120

. By definition

0! = 1

Q2Write

(\binom{n}{r})

using factorials.

(\binom{n}{r}) = \frac{n!}{r! (n - r)!}

. The

(n - r)!

is the part candidates most often forget.

Q3What is

(\binom{8}{3})

56.

\frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56

, and by symmetry it is also

(\binom{8}{5})

Q4What is the nth term of an arithmetic progression?

u_{n} = a + (n - 1) d

, with first term

a

and common difference

d

Q5What is the sum of the first

n

terms of an arithmetic progression?

S_{n} = \frac{n}{2} {2 a + (n - 1) d}

, or

S_{n} = \frac{n}{2} (a + l)

when the last term

l

is known.

Q6When are

a

b

c

in arithmetic progression?

When

2 b = a + c

: the middle term is the average of the other two.

Q7What is the nth term of a geometric progression?

u_{n} = a r^{n - 1}

, with first term

a

and common ratio

r

. The index is

n - 1

, not

n

Q8What is the sum of the first

n

terms of a geometric progression?

S_{n} = \frac{a (r^{n} - 1)}{r - 1}

, which is the same as

\frac{a (1 - r^{n})}{1 - r}

. It needs

r \neq 1

Q9When are

a

b

c

in geometric progression?

When

b^{2} = a c

: the middle term is the geometric mean of the other two.

Q10When does a geometric progression have a sum to infinity?

When

| r | < 1

, so that the terms shrink towards zero. If

| r | \geq 1

the progression diverges.

Q11What is the sum to infinity of a convergent geometric progression?

S_{\infty} = \frac{a}{1 - r}

, used only after checking that

| r | < 1

Q12What is the coefficient of

x^{3}

in the expansion of

(1 + 2 x)^{5}

80. The term is

(\binom{5}{3}) (2 x)^{3} = 10 \times 8 x^{3} = 80 x^{3}

Q13How do you find the common ratio when two terms of a geometric progression are given?

Divide the later term by the earlier one. The

a

cancels and you are left with

r

raised to the number of steps between them.

Q14What is the sum of the first 20 terms of 3, 7, 11, 15, ...?

820.

a = 3

and

d = 4

, so

S_{20} = \frac{20}{2} {6 + 19 (4)} = 10 \times 82

Q15What is the first step in a 'how many terms are needed' question?

Set the

S_{n}

formula equal to the given sum, form a quadratic in

n

, solve it, and reject the root that is not a positive whole number.

0 of 15 marked as known

Exam technique for this topic

Write the formula you are using before you substitute into it. Scheme points are given for $u_{n} = a + (n - 1) d$ or $S_{n} = \frac{n}{2} {2 a + (n - 1) d}$ correctly applied, and those marks survive an arithmetic slip further down.
When two terms are given, subtract for an arithmetic progression and divide for a geometric one. Dividing cancels the unknown first term, which is what makes a geometric question quick, and subtracting does the same job for an arithmetic one.
In a 'how many terms' question you will usually get a quadratic in $n$ . Solve it, then discard the root that is not a positive whole number and say why. Both the quadratic and the rejection are worth marks.
Check $| r | < 1$ in writing before you use $S_{\infty} = \frac{a}{1 - r}$ . A question that asks whether a progression converges wants that condition stated, and a question that asks for a sum to infinity of a progression with $| r | \geq 1$ is looking for the answer that no such sum exists.
Keep the words term and sum apart. 'The amount in the 10th year' is $u_{10}$ ; 'the total after 10 years' is $S_{10}$ . Writing the wrong one down at the start loses every mark that follows.

Why students lose marks here

A student wroteThe nth term of a geometric progression is

a r^{n}

Why it loses marksThe index is one less than the position, because the first term is

a

itself and no multiplication by

r

has happened yet. Using

a r^{n}

gives the term after the one you want.

The answer that scores

u_{n} = a r^{n - 1}

. For 3, 6, 12, ... the 4th term is

3 \times 2^{3} = 24

, not

3 \times 2^{4} = 48

A student wroteThe nth term of an arithmetic progression is

a + n d

Why it loses marksTo reach the nth term you add

d

one time fewer than

n

. Writing

a + n d

gives the

(n + 1)

th term.

The answer that scores

u_{n} = a + (n - 1) d

. For 7, 12, 17, ... the 20th term is

7 + 19 (5) = 102

A student wrote

S_{\infty} = \frac{a}{1 - r}

can be used for any geometric progression.

Why it loses marksThe formula comes from letting

r^{n} \to 0

, which only happens when

| r | < 1

. For

r = 2

r = - 1

the terms do not shrink and the sum has no limit, so the formula produces a meaningless number.

The answer that scoresUse it only when

| r | < 1

. For 5, 10, 20, 40, ... the progression diverges, so there is no sum to infinity.

A student wrote

(\binom{6}{2}) = \frac{6!}{2!} = 360

Why it loses marksThe

(n - r)!

in the denominator has been dropped.

(\binom{n}{r})

counts selections, not arrangements, and dividing only by

r!

counts arrangements instead.

The answer that scores

(\binom{6}{2}) = \frac{6!}{2! 4!} = \frac{720}{2 \times 24} = 15

A student wroteIn

(1 + 3 x)^{6}

the coefficient of

x^{2}

is 9, because

3^{2} = 9

Why it loses marksThe binomial coefficient has been left out. The term in

x^{2}

(\binom{6}{2}) (3 x)^{2}

, so the 15 and the 9 both belong in the answer.

The answer that scores

(\binom{6}{2}) (3 x)^{2} = 15 \times 9 x^{2} = 135 x^{2}

, so the coefficient is 135.

A student wroteFor a geometric progression with

u_{3} = 12

and

u_{6} = 96

, the common ratio is

96 \div 12 = 8

Why it loses marksThree steps of the progression separate the two terms, so the ratio between them is

r^{3}

, not

r

. The cube root is the common ratio, not the quotient.

The answer that scores

\frac{u_{6}}{u_{3}} = r^{3} = 8

, so

r = 2

. Then

a = \frac{12}{2^{2}} = 3

Syllabus coverage

Factorials and binomial coefficients

Expanding $(a + b)^{n}$

Arithmetic progressions

Geometric progressions

Convergence and the sum to infinity

Tackling a series problem

Key terms you must be able to define

Applied

Flashcards

Exam-style questions

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Exam technique for this topic

Why students lose marks here

Series

Syllabus coverage

Factorials and binomial coefficients

Expanding (a+b)n

Arithmetic progressions

Geometric progressions

Convergence and the sum to infinity

Tackling a series problem

Key terms you must be able to define

Applied

Flashcards

Exam-style questions

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Exam technique for this topic

Why students lose marks here

Expanding $(a + b)^{n}$