Add Maths · 12 Series

12 Series

Before you start

Index laws, including negative indices: $\frac{3}{x} = 3 x^{- 1}$ and ${(2 x^{2})}^{3} = 8 x^{6}$ .
Expanding brackets and collecting like terms.
Solving a pair of linear simultaneous equations by elimination or substitution.
Solving a quadratic equation by factorising or by the quadratic formula (which is on the List of formulas).
Using logarithms to solve an equation or inequality with the unknown in a power, such as $2^{k} > 50$ (topic 6).
Modulus notation: $| r | < 1$ means $- 1 < r < 1$ .

Syllabus objectives covered (5)

Use the binomial theorem for expansion of (a + b)ⁿ for positive integer n.
Use the general term \binom{n}{r}aⁿ⁻ʳbʳ, 0 ⩽ r ⩽ n.
Recognise arithmetic and geometric progressions and understand the difference between them.
Use the formulas for the nth 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.

Part 1 of 8

Binomial coefficients: what the numbers count

${(a + b)}^{n}$ means $n$ copies of $(a + b)$ multiplied together. When you expand, every term is made by picking either $a$ or $b$ from each bracket and multiplying what you picked.
If you pick $b$ from $r$ of the brackets, you must pick $a$ from the other $n - r$ , so that term is $a^{n - r} b^{r}$ . The number of ways of choosing which $r$ brackets supply the $b$ is written $(\binom{n}{r})$ , read 'n choose r'. That count is the coefficient of the term.
Check it on ${(a + b)}^{2} = (a + b) (a + b)$ . The term $a b$ arises in two ways ( $a$ from the first bracket and $b$ from the second, or the other way round), and $(\binom{2}{1}) = 2$ . That is where the $2$ in $a^{2} + 2 a b + b^{2}$ comes from.
The formula uses factorials, where $n! = n \times (n - 1) \times \dots \times 2 \times 1$ and $0! = 1$ .
$(\binom{n}{r}) = \frac{n!}{r! (n - r)!}$
Paper 1 has no calculator, so cancel before you multiply. Most of $n!$ cancels with $(n - r)!$ , which leaves $r$ factors counting down from $n$ on top and $r!$ underneath.
$(\binom{8}{3}) = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$
A few values save time: $(\binom{n}{0}) = (\binom{n}{n}) = 1$ and $(\binom{n}{1}) = n$ . Also $(\binom{n}{r}) = (\binom{n}{n - r})$ , because choosing the $r$ brackets that give $b$ is the same as choosing the $n - r$ that give $a$ . So $(\binom{9}{6}) = (\binom{9}{3}) = \frac{9 \times 8 \times 7}{6} = 84$ , which is far quicker by hand.
For small $n$ , Pascal's triangle gives the same numbers. Each row starts and ends with $1$ , and each entry in between is the sum of the two entries just above it. The row $1 4 6 4 1$ gives the coefficients of ${(a + b)}^{4}$ , and the next row, $1 5 10 10 5 1$ , gives those of ${(a + b)}^{5}$ . For larger $n$ the formula is quicker than building rows.
On Paper 2 your calculator may have an $^{n} C_{r}$ key, which gives the same value. Use it to check, but practise by hand, because this topic can be set on Paper 1.

Part 2 of 8

Expanding (a + b)ⁿ in full

The binomial theorem is on the List of formulas. Read it as a pattern rather than as a string of symbols.
${(a + b)}^{n} = a^{n} + (\binom{n}{1}) a^{n - 1} b + (\binom{n}{2}) a^{n - 2} b^{2} + \dots + (\binom{n}{r}) a^{n - r} b^{r} + \dots + b^{n}$
The power of $a$ falls by one each term, from $n$ down to $0$ , while the power of $b$ rises from $0$ to $n$ . In every term the two powers add to $n$ , and there are $n + 1$ terms.
In practice $a$ and $b$ are expressions, not single letters. In ${(2 - 3 x)}^{4}$ , $a = 2$ and $b = - 3 x$ . Put a bracket round each before raising it to a power, because the whole of $b$ , its number and its sign, is raised, not just the $x$ .
Write out the five terms with the coefficients $1, 4, 6, 4, 1$ from Pascal's triangle.
${(2)}^{4} + 4 {(2)}^{3} (- 3 x) + 6 {(2)}^{2} {(- 3 x)}^{2} + 4 (2) {(- 3 x)}^{3} + {(- 3 x)}^{4}$
Now simplify each coefficient to a single number: $4 \times 8 \times (- 3) = - 96$ , $6 \times 4 \times 9 = 216$ and $4 \times 2 \times (- 27) = - 216$ .
${(2 - 3 x)}^{4} = 16 - 96 x + 216 x^{2} - 216 x^{3} + 81 x^{4}$
The signs alternate because $b$ is negative: an odd power of $- 3 x$ is negative and an even power is positive. If $b$ is negative and your signs do not alternate, look for the slip before you go on.
'In ascending powers of $x$ ' means start with the constant term, then the $x$ term, then $x^{2}$ . If a question asks for only the first three terms, you need only $r = 0, 1, 2$ . Do not expand the whole bracket.
'Simplest form' means each coefficient is one number. A term left as $6 \times 4 \times 9 x^{2}$ is not simplified; $216 x^{2}$ is.

Part 3 of 8

The general term: one term without the whole expansion

The general term lets you go straight to any single term of the expansion.
$(\binom{n}{r}) a^{n - r} b^{r}, 0 ⩽ r ⩽ n$
Choose $r$ from the power of $b$ you want: it is the term containing $b^{r}$ . Because the first term, $a^{n}$ , has $r = 0$ , the term containing $b^{r}$ is the $(r + 1)$ th term. So the 4th term has $r = 3$ , not $r = 4$ .
Find the coefficient of $x^{3}$ in ${(3 - 2 x)}^{6}$ . Here $a = 3$ and $b = - 2 x$ , and you want $b^{3}$ , so $r = 3$ .
$(\binom{6}{3}) {(3)}^{3} {(- 2 x)}^{3} = 20 \times 27 \times (- 8 x^{3}) = - 4320 x^{3}$
The term is $- 4320 x^{3}$ and the coefficient is $- 4320$ . Give what the question asks for: a coefficient is the number only, with its sign.
A second bracket in front, as in $(1 + 4 x) {(1 - x)}^{5}$ , needs no new method. To find the coefficient of $x^{2}$ , list every way of making $x^{2}$ : the $1$ times the $x^{2}$ term of ${(1 - x)}^{5}$ , and the $4 x$ times its $x$ term.
From ${(1 - x)}^{5} = 1 - 5 x + 10 x^{2} - \dots$ , the coefficient of $x^{2}$ in the product is $1 \times 10 + 4 \times (- 5) = - 10$ . You only need the terms of the expansion up to the power you are looking for.

Part 4 of 8

When both parts contain x: the term independent of x

In ${(x + \frac{3}{x})}^{4}$ both $a$ and $b$ contain $x$ , so the power of $x$ in a term depends on $r$ in two places. Write $\frac{3}{x}$ as $3 x^{- 1}$ so that you can use the index laws.
Write the general term, then collect every power of $x$ into a single index.
$(\binom{4}{r}) x^{4 - r} {(3 x^{- 1})}^{r} = (\binom{4}{r}) 3^{r} x^{4 - r - r} = (\binom{4}{r}) 3^{r} x^{4 - 2 r}$
'Independent of $x$ ' means the term has no $x$ in it, so the power of $x$ is $0$ . Solve $4 - 2 r = 0$ , which gives $r = 2$ .
Put $r = 2$ back into the general term: $(\binom{4}{2}) \times 3^{2} = 6 \times 9 = 54$ . Check the $x$ really cancels: $x^{2} \times {(3 x^{- 1})}^{2} = x^{2} \times 9 x^{- 2} = 9$ .
The same equation finds any other power. For the coefficient of $x^{2}$ , solve $4 - 2 r = 2$ , so $r = 1$ , and the coefficient is $(\binom{4}{1}) \times 3 = 12$ .
If the equation gives an $r$ that is not a whole number from $0$ to $n$ , there is no such term. In ${(x + \frac{3}{x})}^{5}$ the power is $5 - 2 r$ , which is odd for every whole number $r$ , so it is never $0$ and there is no term independent of $x$ .

Part 5 of 8

Arithmetic or geometric: the test

In an arithmetic progression (AP) you get each term by adding the same number, the common difference $d$ . The sequence $5, 8, 11, 14, \dots$ has $d = 3$ , and $20, 17.5, 15, \dots$ has $d = - 2.5$ .
In a geometric progression (GP) you get each term by multiplying by the same number, the common ratio $r$ . The sequence $3, 6, 12, 24, \dots$ has $r = 2$ , and $40, - 20, 10, - 5, \dots$ has $r = - \frac{1}{2}$ .
To test a sequence, work out the differences $u_{2} - u_{1}, u_{3} - u_{2}, \dots$ and the ratios $\frac{u_{2}}{u_{1}}, \frac{u_{3}}{u_{2}}, \dots$ . All the differences equal means an AP; all the ratios equal means a GP. Check every gap you are given, not just the first. The sequence $2, 4, 8, 14$ starts like a GP, but $\frac{14}{8} \neq 2$ , and its differences $2, 4, 6$ are not equal either, so it is neither.
The two kinds behave differently. An AP changes by the same amount every step, so its terms plotted against $n$ lie on a straight line. A GP changes by the same factor: if $r > 1$ the size of the terms grows faster and faster, if $0 < r < 1$ the terms shrink towards $0$ , and if $r < 0$ they alternate in sign.
graph diagram — not drawn yet
Axes with n from 0 to 7 on the horizontal axis and term value from 0 to 35 on the vertical axis. Plot as separate dots, not joined, the AP u_n = 3n − 1 for n = 1 to 6 (2, 5, 8, 11, 14, 17) and the GP u_n = 2^(n−1) for n = 1 to 6 (1, 2, 4, 8, 16, 32), with a different marker for each and a key; draw a faint dashed straight line through the AP dots only.
The AP dots lie on a straight line because the same amount is added each time. The GP dots curve upwards because each one is double the one before.
Three consecutive terms $p, q, s$ give you an equation. In an AP the two differences are equal, $q - p = s - q$ , so $2 q = p + s$ . In a GP the two ratios are equal, $\frac{q}{p} = \frac{s}{q}$ , so $q^{2} = p s$ .
The two conditions give different answers, which is why you must know which kind you have. If $4, x, 9$ is an AP, $2 x = 13$ and $x = 6.5$ . If $4, x, 9$ is a GP, $x^{2} = 36$ , so $x = 6$ or $x = - 6$ . Both work: $4, - 6, 9$ has ratio $- \frac{3}{2}$ each time.

Part 6 of 8

Arithmetic progressions: the nth term and the sum

Use $a$ for the first term, $d$ for the common difference, $n$ for the number of terms and $l$ for the last term. The formulas are on the List of formulas. What you need is to know where each comes from, so that you pick the right one.
To reach the $n$ th term from the first you add $d$ only $n - 1$ times, because the first term needs no steps. So the 10th term is $a + 9 d$ , not $a + 10 d$ .
$u_{n} = a + (n - 1) d$
For the sum, write the terms forwards and then backwards underneath, and add the two lines in pairs. Every pair adds to $a + l$ , and there are $n$ pairs, so twice the sum is $n (a + l)$ .
$S_{n} = \frac{1}{2} n (a + l) = \frac{1}{2} n [2 a + (n - 1) d]$
The second form comes from replacing $l$ by $a + (n - 1) d$ . Use the first when you know the last term, and the second when you know $d$ .
Two facts about an AP give two equations in $a$ and $d$ . Write each fact with a formula, then solve them simultaneously. 'The 3rd term is $11$ ' gives $a + 2 d = 11$ , and 'the 8th term is $26$ ' gives $a + 7 d = 26$ . Subtracting gives $5 d = 15$ , so $d = 3$ and $a = 5$ .
When the unknown is $n$ , the sum formula gives a quadratic in $n$ . The number of terms must be a positive whole number, so reject a negative root. For 'the least $n$ for which $S_{n}$ exceeds $1000$ ', solve the equation, round up to the next whole number, and check the sums on either side.

Part 7 of 8

Geometric progressions: the nth term and the sum

To reach the $n$ th term you multiply the first term $a$ by $r$ exactly $n - 1$ times. Only $r$ is raised to the power: with $a = 3$ and $r = 2$ , the 5th term is $3 \times 2^{4} = 48$ .
$u_{n} = a r^{n - 1}$
The sum formula comes from a trick worth seeing once. Write out the sum, then multiply every term by $r$ , which shifts each term one place along.
$S_{n} = a + a r + a r^{2} + \dots + a r^{n - 1}, r S_{n} = a r + a r^{2} + \dots + a r^{n - 1} + a r^{n}$
Subtract the second line from the first. Every term cancels except the first term of the top line and the last term of the bottom line.
$S_{n} - r S_{n} = a - a r^{n}$
Factorise both sides and divide by $1 - r$ , which is allowed only when $r \neq 1$ . This is the formula on the List of formulas.
$S_{n} = \frac{a (1 - r^{n})}{1 - r}$
When $r > 1$ , multiplying the top and bottom by $- 1$ gives the same value with fewer negatives: $S_{n} = \frac{a (r^{n} - 1)}{r - 1}$ . For $3 + 6 + 12 + \dots$ to 8 terms, $S_{8} = \frac{3 (2^{8} - 1)}{2 - 1} = 765$ .
Two facts about a GP give two equations such as $a r = 6$ and $a r^{4} = 48$ . Divide one by the other, because $a$ then cancels: $r^{3} = 8$ , so $r = 2$ , and then $a = 3$ .
When the unknown $n$ is in a power, use logarithms. For the first term of $3, 6, 12, \dots$ above $1000$ , solve $3 \times 2^{n - 1} > 1000$ . Then $2^{n - 1} > \frac{1000}{3}$ , so $(n - 1) \lg 2 > \lg \frac{1000}{3}$ , which gives $n - 1 > 8.38 \dots$ and $n > 9.38 \dots$ . The 10th term, $3 \times 2^{9} = 1536$ , is the first above $1000$ ; the 9th is $3 \times 2^{8} = 768$ .
Dividing by $\lg 2$ keeps the inequality the same way round because $\lg 2 > 0$ . If you divide by the logarithm of a number between $0$ and $1$ , which is negative, you must reverse the inequality sign.

Part 8 of 8

Convergence and the sum to infinity

Add the terms of $8 + 4 + 2 + 1 + \dots$ one at a time: the sums are $8, 12, 14, 15, 15.5, 15.75, \dots$ . Each new term covers half of the gap that is left to $16$ , so the sums get as close to $16$ as you like but never reach it. That limit, $16$ , is the sum to infinity.
graph diagram — not drawn yet
Axes with n from 0 to 9 on the horizontal axis and S_n from 0 to 18 on the vertical axis. Plot as separate dots the sums of the first n terms of 8 + 4 + 2 + 1 + … for n = 1 to 8 (8, 12, 14, 15, 15.5, 15.75, 15.875, 15.9375) and draw a dashed horizontal line at S_n = 16 labelled 'S∞ = 16'.
The sums climb towards $16$ and get closer with every term, but never reach or pass it.
Here is why. When $- 1 < r < 1$ , $r^{n}$ gets closer and closer to $0$ as $n$ grows; for example, ${(\frac{1}{2})}^{10} = \frac{1}{1024}$ . So the $r^{n}$ in the sum formula fades away, and $S_{n}$ approaches a fixed value.
$S_{n} = \frac{a (1 - r^{n})}{1 - r} ⟶ \frac{a}{1 - r} = S_{\infty}$
The formula for $S_{\infty}$ is on the List of formulas. It is valid only when the GP is convergent, and applying the condition is your job.
$S_{\infty} = \frac{a}{1 - r} only when | r | < 1, that is - 1 < r < 1$
Check it on $8 + 4 + 2 + \dots$ : $a = 8$ and $r = \frac{1}{2}$ , so $S_{\infty} = \frac{8}{1 - \frac{1}{2}} = 16$ , which matches the sums above.
If $r ⩾ 1$ or $r ⩽ - 1$ , $r^{n}$ does not shrink, so the terms do not shrink and the sums never settle. The sums of $1 + 2 + 4 + \dots$ grow without limit, and the sums of $1 - 1 + 1 - 1 + \dots$ jump $1, 0, 1, 0, \dots$ for ever. Neither has a sum to infinity.
Using the formula anyway gives nonsense. For $3 + 6 + 12 + \dots$ , $\frac{3}{1 - 2} = - 3$ : a negative 'sum' of positive terms. Always find $r$ and check the condition first.
To explain whether a GP has a sum to infinity, quote $r$ and the condition together: ' $r = - \frac{1}{3}$ and $| r | < 1$ , so the progression is convergent and has a sum to infinity.' A sentence without the value of $r$ does not show the reason.
This is another difference between the two kinds of progression. An AP with a non-zero common difference never has a sum to infinity: its terms eventually keep getting larger in size, so its sums grow without limit.
If the ratio contains $x$ , the condition becomes an inequality in $x$ . For $1 + 2 x + 4 x^{2} + \dots$ , $r = 2 x$ , so the progression converges when $- 1 < 2 x < 1$ , that is $- \frac{1}{2} < x < \frac{1}{2}$ . Its sum to infinity is then $\frac{1}{1 - 2 x}$ .

Worked examples

Example 1. (a) Find the first three terms, in ascending powers of $x$ , of the expansion of ${(3 - 2 x)}^{5}$ . (b) Hence find the coefficient of $x^{2}$ in the expansion of $(1 + x) {(3 - 2 x)}^{5}$ .

Show the working

Identify the parts: $a = 3$ , $b = - 2 x$ and $n = 5$ . The first three terms in ascending powers of $x$ are $r = 0, 1, 2$ .
$r = 0$ : the first term is $3^{5} = 243$ .
$r = 1$ : keep $- 2 x$ in a bracket so the sign goes with it.
$(\binom{5}{1}) {(3)}^{4} (- 2 x) = 5 \times 81 \times (- 2 x) = - 810 x$
$r = 2$ : the whole of $- 2 x$ is squared, giving $+ 4 x^{2}$ .
$(\binom{5}{2}) {(3)}^{3} {(- 2 x)}^{2} = 10 \times 27 \times 4 x^{2} = 1080 x^{2}$
For (b), write the product with only the terms you need: $(1 + x) (243 - 810 x + 1080 x^{2} + \dots)$ . An $x^{2}$ term comes from $1 \times 1080 x^{2}$ and from $x \times (- 810 x)$ .
Add the two contributions: $1080 + (- 810) = 270$ . Leaving out the $x \times (- 810 x)$ product would give $1080$ , which is wrong.

Answer: (a) $243 - 810 x + 1080 x^{2}$ . (b) $270$ .

Example 2. Find the term independent of $x$ in the expansion of ${(2 x^{2} - \frac{1}{x})}^{9}$ .

Show the working

Both parts contain $x$ , so use the general term. Write $- \frac{1}{x}$ as $- x^{- 1}$ .
$(\binom{9}{r}) {(2 x^{2})}^{9 - r} {(- x^{- 1})}^{r}$
Collect the powers of $x$ : ${(x^{2})}^{9 - r}$ gives $x^{18 - 2 r}$ and ${(x^{- 1})}^{r}$ gives $x^{- r}$ , so the power of $x$ is $18 - 3 r$ .
Independent of $x$ means the power is $0$ : $18 - 3 r = 0$ , so $r = 6$ . That is a whole number between $0$ and $9$ , so the term exists.
Substitute $r = 6$ . Use $(\binom{9}{6}) = (\binom{9}{3}) = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$ , and ${(- 1)}^{6} = 1$ because the power is even.
$(\binom{9}{6}) {(2 x^{2})}^{3} {(- x^{- 1})}^{6} = 84 \times 8 x^{6} \times x^{- 6} = 672$

Answer: $672$

Example 3. The fourth term of an arithmetic progression is $17$ and the sum of the first $10$ terms is $215$ . (a) Find the first term and the common difference. (b) Find the least number of terms for which the sum of the progression exceeds $1000$ . (A calculator is allowed.)

Show the working

Write the first fact with the $n$ th term formula: the 4th term is $a + 3 d$ .
$a + 3 d = 17$
Write the second fact with the sum formula, $n = 10$ : $5 (2 a + 9 d) = 215$ .
$2 a + 9 d = 43$
Double the first equation to get $2 a + 6 d = 34$ , then subtract it from the second: $3 d = 9$ , so $d = 3$ . Then $a = 17 - 9 = 8$ .
For (b), write $S_{n}$ with these values and simplify.
$S_{n} = \frac{1}{2} n [16 + 3 (n - 1)] = \frac{1}{2} n (3 n + 13)$
Set the sum above $1000$ and multiply by $2$ .
$3 n^{2} + 13 n - 2000 > 0$
Solve $3 n^{2} + 13 n - 2000 = 0$ with the quadratic formula: $n = \frac{- 13 + \sqrt{24169}}{6} \approx 23.7$ . The other root is negative, and a number of terms cannot be negative.
$n$ must be a whole number greater than $23.7$ , so $n = 24$ . Check both sides: $S_{23} = \frac{1}{2} \times 23 \times 82 = 943$ and $S_{24} = \frac{1}{2} \times 24 \times 85 = 1020$ .

Answer: (a) $a = 8$ , $d = 3$ . (b) $24$ terms.

Example 4. The second term of a geometric progression is $- 12$ and the fifth term is $\frac{3}{2}$ . (a) Find the common ratio and the first term. (b) Find the sum of the first $6$ terms. (c) Explain why the progression has a sum to infinity, and find it.

Show the working

Write both facts with $u_{n} = a r^{n - 1}$ : $a r = - 12$ and $a r^{4} = \frac{3}{2}$ .
Divide the second equation by the first so that $a$ cancels.
$r^{3} = \frac{\frac{3}{2}}{- 12} = - \frac{1}{8}$
A real number has exactly one real cube root, and the cube root of a negative number is negative, so $r = - \frac{1}{2}$ . Then $a = \frac{- 12}{- \frac{1}{2}} = 24$ .
Check: the terms are $24, - 12, 6, - 3, \frac{3}{2}$ , so the 2nd and 5th terms are right.
For (b), use the sum formula with $r = - \frac{1}{2}$ . Note ${(- \frac{1}{2})}^{6} = \frac{1}{64}$ because the power is even.
$S_{6} = \frac{24 (1 - \frac{1}{64})}{1 + \frac{1}{2}} = 16 \times \frac{63}{64} = \frac{63}{4}$
For (c), quote $r$ and the condition: $r = - \frac{1}{2}$ and $| r | < 1$ , so the progression is convergent.
Then use the sum to infinity formula.
$S_{\infty} = \frac{24}{1 - (- \frac{1}{2})} = \frac{24}{\frac{3}{2}} = 16$

Answer: (a) $r = - \frac{1}{2}$ , $a = 24$ . (b) $S_{6} = \frac{63}{4}$ . (c) $| r | = \frac{1}{2} < 1$ , so it converges; $S_{\infty} = 16$ .

Where people go wrong

Mistake: Raising only the $x$ to the power, for example writing ${(- 3 x)}^{2}$ as $- 3 x^{2}$ or $3 x^{2}$ inside an expansion.

Instead: Raise the whole of $b$ , number and sign included: ${(- 3 x)}^{2} = 9 x^{2}$ and ${(- 3 x)}^{3} = - 27 x^{3}$ .

In ${(a + b)}^{n}$ , $b$ is the whole second term. The bracket round $- 3 x$ is part of $b$ , so every factor in it gets the power.

Mistake: Using $r = 4$ for the 4th term of a binomial expansion.

Instead: The term containing $b^{r}$ is the $(r + 1)$ th term, so the 4th term has $r = 3$ .

Counting starts at $r = 0$ : the first term, $a^{n}$ , is the $r = 0$ term.

Mistake: Taking the $n$ th term of an AP as $a + n d$ .

Instead: $u_{n} = a + (n - 1) d$ . Test any formula with $n = 1$ : it must give $a$ .

The first term is already $a$ , so only $n - 1$ steps of $d$ are needed to reach the $n$ th term.

Mistake: Reading $a r^{n - 1}$ as ${(a r)}^{n - 1}$ , so that the 5th term of $3, 6, 12, \dots$ becomes $6^{4}$ .

Instead: Only $r$ is raised to the power: $3 \times 2^{4} = 48$ .

A power applies only to the symbol directly in front of it unless there is a bracket.

Mistake: Using $\frac{a}{1 - r}$ for any GP, for example getting $- 3$ as the sum to infinity of $3 + 6 + 12 + \dots$ .

Instead: Find $r$ and check $- 1 < r < 1$ first. Here $r = 2$ , so there is no sum to infinity.

The formula depends on $r^{n}$ shrinking to $0$ , which happens only when $| r | < 1$ .

Check yourself

Expand ${(1 + 2 x)}^{4}$ fully, simplifying each coefficient.
Answer
$1 + 4 (2 x) + 6 {(2 x)}^{2} + 4 {(2 x)}^{3} + {(2 x)}^{4} = 1 + 8 x + 24 x^{2} + 32 x^{3} + 16 x^{4}$ .
Find the coefficient of $x^{3}$ in the expansion of ${(2 - x)}^{7}$ .
Answer
$r = 3$ : $(\binom{7}{3}) \times 2^{4} \times {(- 1)}^{3} = 35 \times 16 \times (- 1) = - 560$ .
Find the term independent of $x$ in the expansion of ${(x - \frac{2}{x})}^{6}$ .
Answer
The general term is $(\binom{6}{r}) {(- 2)}^{r} x^{6 - 2 r}$ . Setting $6 - 2 r = 0$ gives $r = 3$ , so the term is $(\binom{6}{3}) {(- 2)}^{3} = 20 \times (- 8) = - 160$ .
State whether each sequence is arithmetic, geometric or neither, and give $d$ or $r$ where there is one. (a) $7, 3, - 1, - 5$ (b) $3, - 6, 12, - 24$ (c) $1, 4, 9, 16$
Answer
(a) Arithmetic, $d = - 4$ . (b) Geometric, $r = - 2$ . (c) Neither: the differences are $3, 5, 7$ and the ratios are $4, \frac{9}{4}, \frac{16}{9}$ .
On day 1 Mia runs 3 km. Each day after that she runs 0.5 km more than the day before. Find the total distance she runs in the first 14 days.
Answer
An AP with $a = 3$ , $d = 0.5$ , $n = 14$ : $S_{14} = \frac{1}{2} \times 14 \times (6 + 13 \times 0.5) = 7 \times 12.5 = 87.5$ , so 87.5 km.
Find the sum of the first 8 terms of the geometric progression $3 + 6 + 12 + \dots$ .
Answer
$a = 3$ , $r = 2$ : $S_{8} = \frac{3 (2^{8} - 1)}{2 - 1} = 3 \times 255 = 765$ .
Determine which of these geometric progressions has a sum to infinity, and find it where it exists. (a) $8 - 4 + 2 - 1 + \dots$ (b) $2 + 3 + 4.5 + \dots$
Answer
(a) $r = - \frac{1}{2}$ and $| r | < 1$ , so it converges: $S_{\infty} = \frac{8}{1 + \frac{1}{2}} = \frac{16}{3}$ . (b) $r = 1.5$ , so $| r | > 1$ and there is no sum to infinity.
Find the set of values of $x$ for which the geometric progression $1 + 3 x + 9 x^{2} + \dots$ has a sum to infinity.
Answer
$r = 3 x$ , so it converges when $- 1 < 3 x < 1$ , that is $- \frac{1}{3} < x < \frac{1}{3}$ .

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1.

(a) Find the first three terms, in ascending powers of $x$ , of the expansion of ${(2 - \frac{x}{2})}^{6}$ . Give each term in its simplest form. 3 marks No calculator

Mark scheme

B1 $64$
B1 $- 96 x$
B1 $60 x^{2}$

(b) The coefficient of $x^{2}$ in the expansion of $(1 + k x) {(2 - \frac{x}{2})}^{6}$ is $12$ . Find the value of the constant $k$ . 2 marks No calculator

Mark scheme

M1 uses both products that give $x^{2}$ and sets them equal to $12$ : $60 - 96 k = 12$ , FT their coefficients from (a)
A1 $k = \frac{1}{2}$ cao

Question 2.

Find the term independent of $x$ in the expansion of ${(x^{2} - \frac{3}{x})}^{6}$ . 4 marks No calculator

Mark scheme

B1 general term $(\binom{6}{r}) {(x^{2})}^{6 - r} {(- \frac{3}{x})}^{r}$ oe, soi
M1 power of $x$ set to zero, $12 - 3 r = 0$ , giving $r = 4$
M1 dep evaluates $(\binom{6}{4}) {(x^{2})}^{2} {(- \frac{3}{x})}^{4}$ , e.g. $15 \times 81$
A1 $1215$ cao

Question 3. A factory makes $400$ chairs in its first week of production. In each week after that, it makes $25$ more chairs than in the week before.

(a) Find the number of chairs the factory makes in week $20$ . 2 marks

Mark scheme

M1 $400 + 19 \times 25$
A1 $875$

(b) Find the least number of complete weeks after which the total number of chairs made is more than $30 000$ . 4 marks

Mark scheme

M1 $\frac{1}{2} n [800 + 25 (n - 1)]$ set equal to, or compared with, $30 000$
A1 $n^{2} + 31 n - 2400 = 0$ oe three-term quadratic
M1 solves their three-term quadratic, e.g. $n = 35.9$ awrt
A1 $36$ cao. Alternative method: after the first M1, evaluating $S_{35} = 28 875$ and $S_{36} = 30 150$ earns M1 A1, then $36$ A1

Question 4. The first three terms of a geometric progression are $k + 4$ , $k$ and $k - 3$ , where $k$ is a constant.

(a) Find the value of $k$ . 3 marks No calculator

Mark scheme

M1 equates ratios, $\frac{k}{k + 4} = \frac{k - 3}{k}$ , or writes $k^{2} = (k + 4) (k - 3)$
A1 $k^{2} = k^{2} + k - 12$ oe
A1 $k = 12$

(b) Explain why this progression has a sum to infinity. 1 mark No calculator

Mark scheme

B1 $r = \frac{12}{16} = \frac{3}{4}$ and $| r | < 1$ (or $- 1 < r < 1$ ) stated

(c) Find the sum to infinity. 1 mark No calculator

Mark scheme

B1 $64$

How this is examined

Building an AP or GP from two given facts, then using it: a multi-part Find: two facts about terms or sums are given, you find $a$ and $d$ (or $a$ and $r$ ), then a further term, a sum, or the least number of terms; this can be set in context Each fact written as an equation with the right formula, such as $a + 3 d = 17$ and $5 (2 a + 9 d) = 215$ , earns method marks; solving them simultaneously (or dividing $a r^{4}$ by $a r$ to cancel $a$ ) earns the values. For a least number of terms, the quadratic or logarithm inequality must be solved and $n$ rounded up to a whole number.
The coefficient of one power of $x$ when a binomial expansion is multiplied by another bracket: a Find of the first three terms in ascending powers of $x$ , then a 'hence' Find with a bracket such as $(1 + k x)$ in front Each term with its coefficient fully simplified and its sign correct (one mark per term), then every product that makes the target power named and added, then the final value or the value of $k$ .
The term independent of $x$ when both parts of the bracket contain $x$ : a Find on a bracket such as ${(x^{2} - \frac{3}{x})}^{6}$ The general term written with the powers of $x$ collected into one index, such as $x^{12 - 3 r}$ ; that index set to $0$ to find $r$ ; then the value of that one term, with its sign.
Deciding whether a GP has a sum to infinity, and finding it: a one-mark Explain followed by a Find, or a Find of the set of values of $x$ for which a GP with $x$ in its ratio converges The value of $r$ stated with the condition, for example ' $r = \frac{3}{4}$ and $| r | < 1$ '; then $\frac{a}{1 - r}$ with the correct $a$ and $r$ . For a ratio in $x$ , the inequality $- 1 < r < 1$ solved for $x$ .
Using the AP or GP condition to find an unknown term: a Find or Show (that) where three consecutive terms such as $k + 4$ , $k$ , $k - 3$ are in AP or GP The defining property written as an equation, $2 q = p + s$ for an AP or $q^{2} = p s$ (equal ratios) for a GP, then solved for the unknown. Writing out the terms afterwards is a useful check, not a marking point.

Words to know

Binomial coefficient: $(\binom{n}{r}) = \frac{n!}{r! (n - r)!}$ , the number of ways of choosing $r$ things from $n$ ; the coefficient of $a^{n - r} b^{r}$ in ${(a + b)}^{n}$ .
General term: $(\binom{n}{r}) a^{n - r} b^{r}$ , the formula for the $(r + 1)$ th term of the expansion of ${(a + b)}^{n}$ .
Coefficient: The number multiplying a power of the variable, sign included: in $- 4320 x^{3}$ it is $- 4320$ .
Term independent of x: The term of an expansion with no $x$ in it, where the power of $x$ is $0$ .
Arithmetic progression (AP): A sequence in which each term is the previous term plus a fixed number, the common difference $d$ .
Geometric progression (GP): A sequence in which each term is the previous term multiplied by a fixed number, the common ratio $r$ .
Convergent: A GP is convergent when $| r | < 1$ : the sums of its terms approach a fixed value.
Sum to infinity: $S_{\infty} = \frac{a}{1 - r}$ , the value the sums of a convergent GP approach as more terms are added.

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