12 Series | IGCSE Additional Maths 0606

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🎯What you need to be able to do

Expand $ (a + b)^{n} $ for a positive integer $n$, simplifying the coefficients.
Use the general term $ \binom{n}{r}a^{n - r}b^{r} $ to find a particular coefficient, including the term independent of $x$.
Recognise arithmetic and geometric progressions and explain the difference.
Use the $n$th term and sum formulas for both, including in context.
State when a geometric progression has a sum to infinity, explain why, and find it.

📚The mathematics

The binomial theorem

\[ (a + b)^{n} = a^{n} + \binom{n}{1}a^{n - 1}b + \binom{n}{2}a^{n - 2}b^{2} + \cdots + \binom{n}{r}a^{n - r}b^{r} + \cdots + b^{n} \]

This is in the List of formulas. The coefficients $ \binom{n}{r} $ are the rows of Pascal’s triangle, which is quicker for small $n$ on Paper 1. Two habits prevent almost every error: put $b$ in a bracket with its sign, so $ \left(-\tfrac{x}{4}\right)^{2} $ is $ +\tfrac{x^{2}}{16} $, and raise the whole bracket to the power, coefficient and all.

Pascal's triangle from row 0 to row 6. Each entry is the sum of the two above it; row 6, 1 6 15 20 15 6 1, is highlighted as the coefficients used for a power of 6. — Row $n$ gives the coefficients of $ (a + b)^{n} $. Row 6 is used in the worked example.

The general term

The term containing $ b^{r} $ is $ \binom{n}{r}a^{n - r}b^{r} $. To find a specific power of $x$, write the general term, collect the powers of $x$ into one exponent in terms of $r$, and solve for $r$. The term independent of $x$ is the one where that exponent is zero. It is a number: give it as one.

For a product such as $ (1 + 3x)(2 - \tfrac{x}{4})^{6} $, find the few terms of the expansion that can combine to give the power you want, and add the products.

Arithmetic and geometric progressions

AP: add $d$ each time; $ u_{n} = a + (n - 1)d $; $ S_{n} = \tfrac{1}{2}n\{2a + (n - 1)d\} = \tfrac{1}{2}n(a + l) $

GP: multiply by $r$ each time; $ u_{n} = ar^{n - 1} $; $ S_{n} = \dfrac{a(1 - r^{n})}{1 - r} $

These are all in the List of formulas. To show a sequence is an AP, show the differences are equal; for a GP, show the ratios are. For “find the least $n$” with a GP, the inequality ends in $ r^{n} < \ldots $; take logs, and remember that dividing by $ \ln r $ (negative when $ 0 < r < 1 $) reverses the inequality.

Sum to infinity

A GP converges only if $ |r| < 1 $: then $ r^{n} \to 0 $ and

\[ S_{\infty} = \frac{a}{1 - r}, \qquad |r| < 1 \]

To explain why a GP has no sum to infinity, state the ratio and that its modulus is not less than 1. An AP never has a sum to infinity (unless every term is zero).

Partial sums of the geometric progression with first term 24 and ratio three quarters, plotted for n = 1 to 20. They rise quickly, then level off towards the dashed line at 96; S15 = 94.7 is below 95 and S16 = 95.04 is the first above it. — $ a = 24,\ r = \tfrac{3}{4} $: the partial sums approach $ S_{\infty} = 96 $ but never reach it. $ S_{16} $ is the first to exceed 95.

✏️Worked example

(a) Find the first three terms, in ascending powers of $x$, of $ \left(2 - \tfrac{x}{4}\right)^{6} $. [3] (b) Hence find the coefficient of $ x^{2} $ in $ (1 + 3x)\left(2 - \tfrac{x}{4}\right)^{6} $. [2] (c) Find the term independent of $x$ in $ \left(x^{2} - \tfrac{2}{x}\right)^{9} $. [3] (d) A geometric progression has first term 24 and sum to infinity 96. Find the common ratio, and the least number of terms whose sum exceeds 95. [5]

(a) $ 2^{6} + 6(2^{5})\left(-\tfrac{x}{4}\right) + 15(2^{4})\left(-\tfrac{x}{4}\right)^{2} = 64 - 48x + 15x^{2} $.

(b) $ x^{2} $ terms come from $ 1 \times 15x^{2} $ and $ 3x \times (-48x) $: $ 15 - 144 = -129 $.

(c) General term: $ \binom{9}{r}(x^{2})^{9 - r}\left(-\tfrac{2}{x}\right)^{r} = \binom{9}{r}(-2)^{r}x^{18 - 3r} $. Independent of $x$ when $ 18 - 3r = 0 $, $ r = 6 $: $ \binom{9}{6}(-2)^{6} = 84 \times 64 = 5376 $.

(d) $ \dfrac{24}{1 - r} = 96 \Rightarrow 1 - r = \tfrac{1}{4} \Rightarrow r = \tfrac{3}{4} $. Then $ S_{n} = 96\left(1 - 0.75^{n}\right) > 95 $ gives $ 0.75^{n} < \tfrac{1}{96} $, so $ n\ln 0.75 < -\ln 96 $ and, dividing by the negative $ \ln 0.75 $,

\[ n > \frac{\ln 96}{\ln \frac{4}{3}} = 15.87\ldots \quad\Longrightarrow\quad n = 16 \]

Check it. $ S_{15} = 94.7 $ (not enough) and $ S_{16} = 95.04 $ (enough) ✓. In (a), put $ x = 0 $: $ 2^{6} = 64 $, the first term ✓.

Losing the sign of $b$. Writing $ 6(2^{5})\left(\tfrac{x}{4}\right) = 48x $ in (a) makes (b) come out as $ 15 + 144 = 159 $. Bracket the second term with its minus sign, every time.

📝Practise

Written in the style of the current Paper 1 (no calculator) and Paper 2 (calculator) questions.

1. (No calculator.) Find the first four terms of $ (1 + 2x)^{5} $ in ascending powers of $x$. [3]

$ 1 + 5(2x) + 10(2x)^{2} + 10(2x)^{3} = 1 + 10x + 40x^{2} + 80x^{3} $.

2. (No calculator.) Find the coefficient of $ x^{3} $ in the expansion of $ (3 - 2x)^{7} $. [2]

$ \binom{7}{3}(3)^{4}(-2)^{3} = 35 \times 81 \times (-8) = -22680 $.

3. (No calculator.) Find the term independent of $x$ in $ \left(2x + \dfrac{1}{x^{2}}\right)^{6} $. [3]

General term $ \binom{6}{r}(2x)^{6 - r}x^{-2r} = \binom{6}{r}2^{6 - r}x^{6 - 3r} $. $ 6 - 3r = 0 \Rightarrow r = 2 $: $ \binom{6}{2}2^{4} = 15 \times 16 = 240 $.

4. (No calculator.) An arithmetic progression has first term 7 and common difference 4. Find the sum of the 10th to the 25th terms inclusive. [3]

$ S_{25} - S_{9} $. $ S_{25} = \tfrac{25}{2}(14 + 24 \times 4) = 1375 $; $ S_{9} = \tfrac{9}{2}(14 + 8 \times 4) = 207 $. Sum $ = 1168 $. (Subtract $ S_{9} $, not $ S_{10} $: the 10th term is included.)

5. (No calculator.) A geometric progression has second term 12 and fifth term 1.5. Find the first term, the common ratio and the sum to infinity. [4]

$ ar = 12 $, $ ar^{4} = 1.5 $. Dividing, $ r^{3} = \tfrac{1}{8} $, $ r = \tfrac{1}{2} $, so $ a = 24 $. $ S_{\infty} = \dfrac{24}{1 - \frac{1}{2}} = 48 $.

6. (No calculator.) Explain why the geometric progression $ 5, -7.5, 11.25, \ldots $ does not have a sum to infinity. [1]

The common ratio is $ -1.5 $, and $ |-1.5| = 1.5 > 1 $, so the terms do not tend to zero and the sum does not converge.

7. (Calculator.) Ari’s starting salary is $30 000. Plan A raises it by 4% each year; Plan B raises it by $1500 each year. Find the total earned over the first 10 years under each plan, and state which is greater. [5]

Plan A is a GP, $ a = 30000 $, $ r = 1.04 $: $ S_{10} = \dfrac{30000(1.04^{10} - 1)}{0.04} = 360\,000 $ (3 s.f.; $360 183). Plan B is an AP, $ a = 30000 $, $ d = 1500 $: $ S_{10} = 5(60000 + 9 \times 1500) = 367\,500 $. Plan B pays more over 10 years (though a percentage rise always wins eventually).

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

Desmos — plot $ S_{n} $ for a GP and watch it level off at $ \tfrac{a}{1 - r} $
Khan Academy — the binomial theorem and arithmetic and geometric series