Mathematics guide, Foundations, step 6 of 33Previous step Next step

Step 6: Sequences, Series and the Binomial Theorem

By the EconTutorials team. 11-minute read with 6 practice questions.

Syllabus9709 P1 1.6, P3 3.1 · IB SL 1.2, 1.3, 1.9

In 30 seconds

An arithmetic progression adds the same $ d $ each time: $ u_n = a + (n-1)d $ and $ S_n = \frac{n}{2}(a + l) = \frac{n}{2}[2a + (n-1)d] $.
A geometric progression multiplies by the same $ r $ each time: $ u_n = ar^{n-1} $ and $ S_n = \dfrac{a(1-r^n)}{1-r} $. Only when $ |r| < 1 $ does it have a sum to infinity, $ S_\infty = \dfrac{a}{1-r} $.
The binomial theorem expands $ (a + b)^n $ for a positive whole number $ n $: the general term is $ \binom{n}{r}a^{n-r}b^r $, where $ \binom{n}{r} $ counts the ways to choose $ r $ brackets.
Exam point: for any other rational $ n $, $ (1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots $ never stops and is valid only for $ |x| < 1 $. For $ (a + bx)^n $, take out $ a^n $ first.

Video lesson coming soonA short animated lesson for this step will appear here.

Sequences and sigma notation

A sequence is a list with a rule. A series is what you get when you add the list up. Two rules matter more than any others: add the same amount each time and multiply by the same amount each time. Between them they price loans, compare pay offers and explain why government spending ripples through an economy.

A rule can be given in two ways. An $ n $th-term formula gives any term directly: if $ u_n = 4n-1 $, then $ u_1 = 3 $, $ u_2 = 7 $ and you can jump straight to $ u_{50} = 199 $. A recurrence relation gives each term from the one before, plus a starting value: if $ u_1 = 4 $ and $ u_{n+1} = 3u_n-5 $, then $ u_2 = 7 $, $ u_3 = 16 $, $ u_4 = 43 $. Recurrences are often easier to write down from a real situation, but to reach $ u_{50} $ you would need every term before it.

Long sums are written with the Greek capital sigma, $ \Sigma $, meaning “add up”:

\[ \sum_{r=1}^{6}(3r-1) = 2 + 5 + 8 + 11 + 14 + 17 = 57 \]

Put $ r = 1, 2, \dots, 6 $ into $ 3r-1 $ in turn and add the results. A sum from $ r = m $ to $ r = n $ has $ n-m + 1 $ terms, not $ n-m $: from $ r = 4 $ to $ r = 15 $ there are 12 terms.

Arithmetic progressions

In an arithmetic progression (AP) you add the same number, the common difference $ d $, each time: $ a,\ a + d,\ a + 2d,\ \dots $ The $ n $th term has had $ d $ added $ n-1 $ times, because the first term has had it added no times at all.

\[ u_n = a + (n-1)d \qquad S_n = \frac{n}{2}(a + l) = \frac{n}{2}\big[2a + (n-1)d\big] \]

Where does the sum come from? Write the sum forwards, then backwards underneath. The first and last terms add to $ a + l $. The second and second-last add to $ (a + d) + (l-d) = a + l $ again, and so does every pair. With $ n $ pairs, twice the sum is $ n(a + l) $. In words: the number of terms times the average of the first and last.

Key termArithmetic progression: a sequence in which $ u_{n+1}-u_n $ is the same constant $ d $ for every $ n $.

Worked example

The 4th term of an arithmetic progression is 23 and the 10th term is 59. Find the first term, the common difference and the sum of the first 20 terms.

Step 1Write each fact with the $ n $th-term formula: $ a + 3d = 23 $ and $ a + 9d = 59 $.

Step 2Subtract: $ 6d = 36 $, so $ d = 6 $. Then $ a = 23-18 = 5 $.

Step 3$ S_{20} = \frac{20}{2}\big[2(5) + 19(6)\big] = 10 \times 124 = 1240 $.

The progression is $ 5, 11, 17, 23, \dots $, and its 20th term is $ 5 + 19 \times 6 = 119 $. The other form gives the same sum: $ 10 \times (5 + 119) = 1240 $.

Worked example

An arithmetic progression has first term 3 and common difference 5. Find the least number of terms for which the sum exceeds 1000.

Step 1$ S_n = \frac{n}{2}\big[6 + 5(n-1)\big] = \frac{n}{2}(5n + 1) $. We need $ \frac{n}{2}(5n + 1) > 1000 $, so $ 5n^2 + n-2000 > 0 $.

Step 2The positive root of $ 5n^2 + n-2000 = 0 $ is $ n = \dfrac{-1 + \sqrt{40001}}{10} = 19.9 $ to 3 significant figures.

Step 3So the least whole number is $ n = 20 $. Check: $ S_{19} = 912 $ and $ S_{20} = 1010 $.

Always check the whole numbers either side of the root. It confirms both the rounding and the direction of the inequality.

Geometric progressions

In a geometric progression (GP) you multiply by the same number, the common ratio $ r $, each time: $ a,\ ar,\ ar^2,\ \dots $ To find $ r $, divide any term by the one before it.

\[ u_n = ar^{n-1} \qquad S_n = \frac{a(1-r^n)}{1-r} = \frac{a(r^n-1)}{r-1} \qquad (r \neq 1) \]

The sum formula has a neat proof. Write $ S_n = a + ar + \dots + ar^{n-1} $, then multiply every term by $ r $: $ rS_n = ar + ar^2 + \dots + ar^n $. The two lines match except at the ends, so subtracting leaves $ S_n-rS_n = a-ar^n $. Factorise and divide by $ 1-r $. (When $ r = 1 $ every term is $ a $, and $ S_n = na $.)

Worked example

The 2nd term of a geometric progression is 6 and the 5th term is 162. Find the first term, the common ratio and the sum of the first 8 terms.

Step 1$ ar = 6 $ and $ ar^4 = 162 $. Divide the second by the first: $ r^3 = 27 $, so $ r = 3 $.

Step 2Then $ a = 6 \div 3 = 2 $.

Step 3$ S_8 = \dfrac{2(3^8-1)}{3-1} = 3^8-1 = 6560 $.

Dividing one equation by the other cancels $ a $. When the unknown is the number of terms, it sits in an index, and you need the logarithms of Step 5.

Common mistakeDividing by a negative logarithm without reversing the inequality. If $ 0.8^n < 0.01 $, then $ n\ln 0.8 < \ln 0.01 $, and $ \ln 0.8 $ is negative. Dividing by it gives $ n > \frac{\ln 0.01}{\ln 0.8} = 20.6 $, so $ n \ge 21 $. Forgetting to reverse gives $ n < 20.6 $, which is nonsense: the terms shrink, so you need more of them, not fewer.

An economics example: a fixed raise or a percentage raise? Two jobs both start at \$30,000. Offer A rises by \$1,800 a year: an AP with $ d = 1800 $. Offer B rises by 5% a year: a GP with $ r = 1.05 $. At first A pays more; in year 2 it is \$31,800 against \$31,500. But an AP adds a fixed amount while a GP multiplies, and multiplying always wins in the end. In year 10, A pays $ 30000 + 9(1800) = \$46{,}200 $ and B pays $ 30000 \times 1.05^9 = \$46{,}539.85 $ (Figure 1). Over the first 10 years, though, A still pays more in total: $ \frac{10}{2}(30000 + 46200) = \$381{,}000 $ against B’s $ \frac{30000(1.05^{10}-1)}{0.05} = \$377{,}336.78 $. Which offer is better depends on how long you plan to stay.

A fixed raise against a percentage raise salary in $

Offer A: +$1,800 a year (arithmetic)Offer B: +5% a year (geometric)

Figure 1. Offer A’s salaries lie on a straight line; offer B’s curve upwards. B first pays more in year 10 ($46,540 against $46,200), and the gap then widens every year.

The sum to infinity

Can infinitely many numbers add up to something finite? Walk half the way to a wall, then half the remaining distance, and so on. You take infinitely many steps but never pass the wall, so $ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots $ should be 1.

The formula shows when this works. In $ S_n = \frac{a(1-r^n)}{1-r} $ only $ r^n $ depends on $ n $. If $ |r| < 1 $, each multiplication by $ r $ makes $ r^n $ smaller in size, so $ r^n \to 0 $ and $ S_n $ approaches $ \frac{a}{1-r} $. When $ |r| \ge 1 $ the terms do not shrink, and the partial sums never settle.

\[ S_\infty = \frac{a}{1-r} \qquad \text{provided } -1 < r < 1 \]

Key termConvergent series: a series whose partial sums $ S_n $ approach a finite limit as $ n \to \infty $. A series that is not convergent is divergent and has no sum to infinity.

Worked example

The geometric series $ 2 + 4x + 8x^2 + \dots $ is convergent. Find the set of possible values of $ x $ and an expression for the sum to infinity. Find $ x $ when the sum to infinity is 5.

Step 1Here $ a = 2 $ and $ r = 2x $. It converges when $ |2x| < 1 $, that is $ -\frac{1}{2} < x < \frac{1}{2} $.

Step 2$ S_\infty = \dfrac{2}{1-2x} $.

Step 3If $ \dfrac{2}{1-2x} = 5 $, then $ 1-2x = \frac{2}{5} $, so $ x = \frac{3}{10} $.

Finish by checking that $ x = \frac{3}{10} $ lies inside $ -\frac{1}{2} < x < \frac{1}{2} $. It does, so the answer is valid.

An economics example: the Keynesian multiplier. The government spends an extra \$1 billion. The people who receive it spend a fraction of it, their marginal propensity to consume (MPC), and save the rest. With MPC = 0.75, \$0.75 billion is spent in shops, whose owners spend 0.75 of that, and so on. The total extra spending, in billions of dollars, is

\[ 1 + 0.75 + 0.75^2 + \dots = \frac{1}{1-0.75} = 4 \]

So the multiplier is $ \frac{1}{1-\text{MPC}} $, the sum to infinity of a GP with ratio MPC. It is finite because $ 0 < \text{MPC} < 1 $. Figure 2 shows the rounds shrinking and the total closing in on 4. The same series values a perpetuity, which pays a fixed sum every year for ever: at 4% interest, \$50 a year is worth $ \frac{50/1.04}{1-1/1.04} = \frac{50}{0.04} = \$1{,}250 $ today.

The multiplier with MPC = 0.75 $ billion

Figure 2. Round k adds 0.75ᵏ⁻¹ billion dollars of spending (bars). The running total (dots) climbs towards 4 but never passes it: after 6 rounds it is already 3.29.

Common mistakeUsing $ \frac{a}{1-r} $ when $ |r| \ge 1 $. “$ 1 + 3 + 9 + \dots = \frac{1}{1-3} = -\frac{1}{2} $” is plainly absurd for a sum of growing positive terms. Check $ |r| < 1 $ and say so before you write $ S_\infty $.

The binomial theorem

Expand $ (a + b)^2 $, $ (a + b)^3 $ and $ (a + b)^4 $ and the coefficients are 1 2 1, then 1 3 3 1, then 1 4 6 4 1. These rows form Pascal’s triangle, where each number is the sum of the two above it; the next row is 1 5 10 10 5 1. In each term the power of $ a $ falls as the power of $ b $ rises, and the two always add to $ n $.

For large $ n $ you want a formula instead of the triangle. Entry $ r $ of row $ n $ (counting from $ r = 0 $) is the binomial coefficient

\[ \binom{n}{r} = \frac{n!}{r!\,(n-r)!} \qquad (a + b)^n = \sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r \]

Why these numbers? $ (a + b)^n $ is $ n $ brackets multiplied together. To make one term you pick $ a $ or $ b $ from each bracket. The term $ a^{n-r}b^r $ appears once for every way of choosing which $ r $ brackets supply a $ b $, and there are $ \binom{n}{r} $ such ways. Also written $ {}^nC_r $, it is found on calculators as nCr: for example $ \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15 $.

Worked example

(a) Find the coefficient of $ x^3 $ in the expansion of $ (2-x)^7 $. (b) Find the term independent of $ x $ in $ \left(x + \dfrac{3}{x^2}\right)^6 $.

Step 1(a) The general term is $ \binom{7}{r}2^{7-r}(-x)^r $. For $ x^3 $, take $ r = 3 $.

Step 2$ \binom{7}{3}\,2^4\,(-x)^3 = 35 \times 16 \times (-x^3) = -560x^3 $. The coefficient is $ -560 $.

Step 3(b) The general term is $ \binom{6}{r}x^{6-r}\left(\frac{3}{x^2}\right)^r = \binom{6}{r}3^r x^{6-3r} $.

Step 4“Independent of $ x $” means $ 6-3r = 0 $, so $ r = 2 $. The term is $ \binom{6}{2}3^2 = 15 \times 9 = 135 $.

Writing the general term first, then simplifying its power of $ x $, turns both questions into one small equation for $ r $.

Common mistakeLosing the sign or the coefficient inside the bracket. In $ (2-x)^7 $ the second term is $ -x $, so its cube is $ (-x)^3 = -x^3 $. In $ (1-3x)^4 $ the term in $ x^2 $ uses $ (-3x)^2 = 9x^2 $, not $ -3x^2 $. Keep the whole of $ b $, sign included, in brackets and raise all of it to the power.

The binomial expansion for any rational power

What is $ \sqrt{1 + x} $ or $ (1 + x)^{-1} $ as a series? Write the binomial coefficients in their product form:

\[ \begin{aligned} (1 + x)^n = 1 &+ nx + \frac{n(n-1)}{2!}x^2 \\ &+ \frac{n(n-1)(n-2)}{3!}x^3 + \dots \qquad |x| < 1 \end{aligned} \]

None of these coefficients needs $ n $ to be a whole number. If $ n $ is a positive integer, the coefficient of $ x^{n+1} $ contains the factor $ n-n = 0 $, so the series stops and is valid for all $ x $. For any other $ n $, no factor is ever zero and the series runs for ever. An infinite series must converge, and that is where $ |x| < 1 $ comes from. For $ n = -1 $ the series is $ 1-x + x^2-x^3 + \dots $, a geometric series with ratio $ -x $, whose sum to infinity is $ \frac{1}{1 + x} $ only when $ |x| < 1 $.

Worked example

Expand $ \sqrt{1-4x} $ in ascending powers of $ x $ up to the term in $ x^3 $, and state the values of $ x $ for which it is valid. Use it to estimate $ \sqrt{0.96} $.

Step 1Use $ n = \frac{1}{2} $ with $ -4x $ in place of $ x $ everywhere: $ 1 + \frac{1}{2}(-4x) + \dfrac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2}(-4x)^2 + \dfrac{\frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{6}(-4x)^3 $.

Step 2Simplify each term: $ 1-2x-\frac{1}{8}(16x^2) + \frac{1}{16}(-64x^3) = 1-2x-2x^2-4x^3 $.

Step 3Valid when $ |-4x| < 1 $, that is $ |x| < \frac{1}{4} $.

Step 4$ \sqrt{0.96} = \sqrt{1-4(0.01)} $, so put $ x = 0.01 $: $ 1-0.02-0.0002-0.000004 = 0.979796 $.

The true value is $ 0.9797959\ldots $, so four terms are correct to 6 decimal places. Figure 3 shows why this works only for small $ x $.

√(1 + x) and its binomial approximations

y = √(1 + x)1 + x/21 + x/2 − x²/8 + x³/16

Figure 3. Inside |x| < 1 (shaded) the partial sums hug the curve, and more terms fit better. Outside it the cubic term takes over and the approximation drifts away.

The formula needs a 1 at the front. For $ (a + bx)^n $, take out $ a^n $ first: $ (a + bx)^n = a^n\left(1 + \frac{b}{a}x\right)^n $, valid for $ |x| < \left|\frac{a}{b}\right| $. The factor you take out is raised to the power $ n $ too, so $ (9 + x)^{1/2} = 3\left(1 + \frac{x}{9}\right)^{1/2} $, not $ 9\left(1 + \frac{x}{9}\right)^{1/2} $.

Exam tipThe binomial theorem for positive whole numbers, APs and GPs are Cambridge 9709 Paper 1 and IB SL content. Rational powers with a validity condition are Paper 3 and IB HL only, and AP Calculus does not test them; AP BC meets infinite series in Unit 10, where the geometric series is the first one studied. Whenever you expand for a non-integer $ n $, state the range of validity: it usually carries a mark.

Practise: 6 questions

Try each one on paper first, then check the worked answer.

EasyQ1. For the arithmetic progression $ 7, 11, 15, \dots $, find the 25th term and the sum of the first 25 terms. Show worked answer

$ a = 7 $, $ d = 4 $.
$ u_{25} = 7 + 24 \times 4 = 103 $.
$ S_{25} = \frac{25}{2}(7 + 103) = \frac{25}{2} \times 110 = 1375 $.

EasyQ2. For the geometric progression $ 48, 24, 12, \dots $, find the 6th term and the sum to infinity. Show worked answer

$ a = 48 $, $ r = \frac{1}{2} $.
$ u_6 = 48 \times \left(\frac{1}{2}\right)^5 = \frac{48}{32} = 1.5 $.
$ |r| < 1 $, so $ S_\infty = \dfrac{48}{1-\frac{1}{2}} = 96 $.

MediumQ3. Find the coefficient of $ x^2 $ in the expansion of $ (3 + 2x)^5 $. Show worked answer

General term: $ \binom{5}{r}3^{5-r}(2x)^r $. Take $ r = 2 $.
$ \binom{5}{2} \times 3^3 \times (2x)^2 = 10 \times 27 \times 4x^2 = 1080x^2 $.
The coefficient is 1080.

MediumQ4. A geometric progression has first term 5 and common ratio 1.2. Find the least number of terms for which the sum exceeds 200. Show worked answer

$ S_n = \dfrac{5(1.2^n-1)}{0.2} = 25(1.2^n-1) $.
$ 25(1.2^n-1) > 200 $, so $ 1.2^n > 9 $.
$ n\ln 1.2 > \ln 9 $, and $ \ln 1.2 > 0 $, so $ n > \dfrac{\ln 9}{\ln 1.2} = 12.05 $.
The least number of terms is 13. Check: $ 1.2^{12} = 8.92 $ and $ 1.2^{13} = 10.7 $.

MediumQ5. (Cambridge P3, IB HL) Expand $ (2 + x)^{-2} $ in ascending powers of $ x $ up to the term in $ x^3 $, and state the set of values of $ x $ for which the expansion is valid. Show worked answer

Take out $ 2^{-2} = \frac{1}{4} $: $ (2 + x)^{-2} = \frac{1}{4}\left(1 + \frac{x}{2}\right)^{-2} $.
$ \left(1 + \frac{x}{2}\right)^{-2} = 1 + (-2)\frac{x}{2} + \frac{(-2)(-3)}{2}\left(\frac{x}{2}\right)^2 + \frac{(-2)(-3)(-4)}{6}\left(\frac{x}{2}\right)^3 = 1-x + \frac{3}{4}x^2-\frac{1}{2}x^3 $.
So $ (2 + x)^{-2} = \frac{1}{4}-\frac{1}{4}x + \frac{3}{16}x^2-\frac{1}{8}x^3 + \dots $
Valid for $ \left|\frac{x}{2}\right| < 1 $, that is $ |x| < 2 $.

ExamQ6. An arithmetic progression has first term $ a $ and common difference $ d $, where $ d \neq 0 $. The 1st, 3rd and 7th terms of the AP are the first three terms of a geometric progression. (a) Show that $ a = 2d $. [3] (b) Find the common ratio of the GP. [2] (c) Given that $ a = 6 $, find the sum of the first 10 terms of the AP and the sum of the first 10 terms of the GP. [3] Show worked answer

(a) The terms are $ a $, $ a + 2d $, $ a + 6d $. For a GP, $ \dfrac{a + 2d}{a} = \dfrac{a + 6d}{a + 2d} $ [M1], so $ (a + 2d)^2 = a(a + 6d) $.
$ a^2 + 4ad + 4d^2 = a^2 + 6ad $, so $ 4d^2 = 2ad $ [A1]. Since $ d \neq 0 $, divide by $ 2d $: $ a = 2d $ [A1].
(b) $ r = \dfrac{a + 2d}{a} = \dfrac{4d}{2d} $ [M1] $ = 2 $ [A1].
(c) $ d = 3 $. AP: $ S_{10} = 5(12 + 9 \times 3) = 195 $ [M1 A1]. The GP is $ 6, 12, 24, \dots $, so its $ S_{10} = 6(2^{10}-1) = 6138 $ [A1].
Total: 8 marks.

Check yourself

1. How many terms are there in $ \displaystyle\sum_{r=4}^{15} r^2 $?

A sum from $ r = m $ to $ r = n $ has $ n-m + 1 $ terms: $ 15-4 + 1 = 12 $.

2. Which common ratio gives a geometric series with a sum to infinity?

A sum to infinity exists only when $ |r| < 1 $. Only $ |-0.9| = 0.9 $ is below 1; the sign does not matter.

3. What is the coefficient of $ x^2 $ in the expansion of $ (1 + x)^{-1} $?

$ \frac{n(n-1)}{2!} = \frac{(-1)(-2)}{2} = 1 $. The series is $ 1-x + x^2-x^3 + \dots $

Your score: 0 out of 3

The short version

A sequence is given by an $ n $th-term formula or a recurrence; $ \sum_{r=m}^{n} $ has $ n-m + 1 $ terms.
AP: $ u_n = a + (n-1)d $ and $ S_n = \frac{n}{2}(a + l) $, from pairing first and last terms.
GP: $ u_n = ar^{n-1} $ and $ S_n = \frac{a(1-r^n)}{1-r} $, from subtracting $ rS_n $. $ S_\infty = \frac{a}{1-r} $ only when $ |r| < 1 $. The multiplier $ \frac{1}{1-\text{MPC}} $ is a sum to infinity.
$ (a + b)^n = \sum \binom{n}{r}a^{n-r}b^r $ for positive whole $ n $. Write the general term first to find one coefficient.
For rational $ n $, $ (1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots $ is infinite and valid for $ |x| < 1 $; for $ (a + bx)^n $, take out $ a^n $ and use $ |x| < \left|\frac{a}{b}\right| $.
The full chapter in Mathematics Made Simple has many more worked examples and a full set of practice questions with a complete answer key.

Step 6: Sequences, Series and the Binomial Theorem

Sequences and sigma notation

Arithmetic progressions

Geometric progressions

The sum to infinity

The binomial theorem

The binomial expansion for any rational power

Practise: 6 questions

Check yourself

The short version

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