More Binomial Expansions

A LevelAQAEdexcelOCR

More Binomial Expansions

In this section we shall look at some advanced skills involving binomial expansion, including partial fractions and approximating.

Make sure you are happy with the following topics before continuing.

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Partial Fractions

We can find the binomial expansion of complicated functions by first decomposing them into partial fractions.

Example: Find the first three terms of the expansion of $dfrac{2x+1}{(x-1)(x+2)}$ .

$begin{aligned}dfrac{2x+1}{(x-1)(x+2)}&=dfrac{1}{x-1}+dfrac{1}{x+2}[1.2em]&=(x-1)^{-1}+(x+2)^{-1}end{aligned}$

Now we can do binomial expansion on $(x-1)^{-1}$ and $(x+2)^{-1}$

$begin{aligned}&dfrac{2x+1}{(x-1)(x+2)}=(-1)^{-1}(1-x)^{-1}+2^{-1}left(1+dfrac{1}{2}xright)^{-1}[1.2em]&=-left(1-(-x)+dfrac{-1times(-2)}{1times2}(-x)^{2}+…right)[1.2em]&+dfrac{1}{2}left(1-dfrac{1}{2}x+dfrac{-1times(-2)}{1times2}left(dfrac{1}{2}xright)^{2}+…right)[1.2em]&=-left(1+x+dfrac{2}{2}x^{2}+…right)[1.2em]&+dfrac{1}{2}left(1-dfrac{1}{2}x+dfrac{2}{2}timesdfrac{1}{4}x^{2}+…right)[1.2em]&=-(1+x+x^{2}+…)+dfrac{1}{2}left(1-dfrac{1}{2}x+dfrac{1}{4}x^{2}+…right)[1.2em]&=-1-x-x^{2}+dfrac{1}{2}-dfrac{1}{4}x+dfrac{1}{8}x^{2}+…[1.2em]&=-dfrac{1}{2}-dfrac{5}{4}x-dfrac{7}{8}x^{2}+…end{aligned}$

Approximations from Binomial Expansions

By substituting in certain values for $x$ , we can use the binomial expansion to approximate things.

Example: Use the binomial expansion of $(1-3x)^{frac{1}{4}}$ to four terms to find $sqrt[4]{0.97}$

$begin{aligned}(1-3x)^{frac{1}{4}}&=1-left(dfrac{1}{4}times3xright)+left(dfrac{dfrac{1}{4}times -dfrac{3}{4}}{1times2}times(-3x)^{2}right)[1.2em]&+left(dfrac{dfrac{1}{4}times -dfrac{3}{4}times-dfrac{7}{4}}{1times2times3}times(-3x)^{3}right)+…[1.2em]&=1-dfrac{3}{4}x+left(dfrac{-dfrac{3}{16}}{2}times9x^{2}right)-left(dfrac{dfrac{21}{64}}{6}times27x^{3}right)+…[1.2em]&=1-dfrac{3}{4}x-dfrac{3times9}{16times2}x^{2}-dfrac{21times27}{64times6}x^{3}[1.2em]&=1-dfrac{3}{4}x-dfrac{27}{32}x^{2}-dfrac{567}{384}x^{3}[1.2em]&=1-dfrac{3}{4}x-dfrac{27}{32}x^{2}-dfrac{189}{128}x^{3}end{aligned}$

Now substitute $x=dfrac{1}{100}$

$left(1-3timesdfrac{1}{100}right)^{frac{1}{4}}=1-left(dfrac{3}{4}timesdfrac{1}{100}right)-left(dfrac{27}{32}timesleft(dfrac{1}{100}right)^{2}right)-left(dfrac{189}{128}timesleft(dfrac{1}{100}right)^{3}right)$

$(1-3times0.01)^{frac{1}{4}}=1-dfrac{3}{400}-left(dfrac{27}{32}timesdfrac{1}{10000}right)-left(dfrac{189}{128}timesdfrac{1}{1000000}right)$

$(1-0.03)^{frac{1}{4}}=1-dfrac{3}{400}-dfrac{27}{320000}-dfrac{189}{128000000}$

$0.97^{frac{1}{4}}=0.9916547734$

So our estimate is $sqrt[4]{0.97}=0.9916547734$ , which is very close to the real value of $0.9924141173$ . Clearly, this approximation method is a very powerful tool.

More Binomial Expansions Example Questions

Question 1: Given that $dfrac{6x^{2}+25x+23}{(x+1)(x+2)(x+3)}=dfrac{2}{x+1}+dfrac{3}{x+2}+dfrac{1}{x+3}$ , find the binomial expansion of $dfrac{6x^{2}+25x+23}{(x+1)(x+2)(x+3)}$ up to and including the $x^{2}$ term.

[6 marks]

$begin{aligned}&dfrac{6x^{2}+25x+23}{(x+1)(x+2)(x+3)}=dfrac{2}{x+1}+dfrac{3}{x+2}+dfrac{1}{x+3}[1.2em]&=2(x+1)^{-1}+3(x+2)^{-1}+(x+3)^{-1}[1.2em]&=2(1+x)^{-1}+3times2^{-1}left(1+dfrac{1}{2}xright)^{-1}[1.2em]&+3^{-1}left(1+dfrac{1}{3}xright)^{-1}[1.2em]&=2(1+x)^{-1}+3timesdfrac{1}{2}left(1+dfrac{1}{2}xright)^{-1}[1.2em]&+dfrac{1}{3}left(1+dfrac{1}{3}xright)^{-1}[1.2em]&=2(1+x)^{-1}+dfrac{3}{2}left(1+dfrac{1}{2}xright)^{-1}+dfrac{1}{3}left(1+dfrac{1}{3}xright)^{-1}[1.2em]&=2left(1-x+dfrac{-1times(-2)}{1times2}x^{2}right)[1.2em]&+dfrac{3}{2}left(1-dfrac{1}{2}x+dfrac{-1times(-2)}{1times2}left(dfrac{1}{2}xright)^{2}right)[1.2em]&+dfrac{1}{3}left(1-dfrac{1}{3}x+dfrac{-1times(-2)}{1times2}left(dfrac{1}{3}xright)^{2}right)[1.2em]&=2left(1-x+dfrac{2}{2}x^{2}right)+dfrac{3}{2}left(1-dfrac{1}{2}x+dfrac{2}{2}timesdfrac{1}{4}x^{2}right)[1.2em]&+dfrac{1}{3}left(1-dfrac{1}{3}x+dfrac{2}{2}timesdfrac{1}{9}x^{2}right)[1.2em]&=2(1-x+x^{2})+dfrac{3}{2}left(1-dfrac{1}{2}x+dfrac{1}{4}x^{2}right)[1.2em]&+dfrac{1}{3}left(1-dfrac{1}{3}x+dfrac{1}{9}x^{2}right)[1.2em]&=2-2x+2x^{2}+dfrac{3}{2}-dfrac{3}{4}x+dfrac{3}{8}x^{2}+dfrac{1}{3}-dfrac{1}{9}x[1.2em]&+dfrac{1}{27}x^{2}[1.2em]&=dfrac{23}{6}-dfrac{103}{36}x+dfrac{521}{216}x^{2}end{aligned}$

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Question 2: Use the approximation $(1-8x)^{frac{1}{3}}=1-dfrac{8}{3}x-dfrac{64}{9}x^{2}+…$ to find $sqrt[3]{0.84}$ , giving your answer to $4$ decimal places.

This expansion is valid for $|x|<dfrac{1}{8}$

[3 marks]

Try $x=0.02$

$(1-8times0.02)^{frac{1}{3}}=1-left(dfrac{8}{3}times0.02right)-left(dfrac{64}{9}times0.02^{2}right)+…$

$(1-0.16)^{frac{1}{3}}=1-dfrac{0.16}{3}-left(dfrac{64}{9}times0.0004right)+…$

$(0.84)^{frac{1}{3}}=1-dfrac{16}{300}-dfrac{0.0256}{9}+…$

$begin{aligned}sqrt[3]{0.84}&=1-dfrac{4}{75}-dfrac{256}{90000}+…[1.2em]&=1-dfrac{4}{75}-dfrac{16}{5625}+…[1.2em]&=1-dfrac{4}{75}-dfrac{16}{5625}+…[1.2em]&=dfrac{5309}{5625}[1.2em]&=0.9438end{aligned}$

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Question 3:

i) Express $dfrac{2+9x}{(1+5x)(1+4x)}$ as partial fractions.

ii) Hence find the first three terms of the binomial expansion of $dfrac{2+9x}{(1+5x)(1+4x)}$

[7 marks]

i) $dfrac{2+9x}{(1+5x)(1+4x)}=dfrac{A}{1+5x}+dfrac{B}{1+4x}$

$2+9x=A(1+4x)+B(1+5x)$

$2+9x=A+4Ax+B+5Bx$

$A+B=2;;5A+4B=9$

$A=1;;B=1$

$dfrac{2+9x}{(1+5x)(1+4x)}=dfrac{1}{1+5x}+dfrac{1}{1+4x}$

ii)

$begin{aligned}&dfrac{2+9x}{(1+5x)(1+4x)}=dfrac{1}{1+5x}+dfrac{1}{1+4x}[1.2em]&=(1+5x)^{-1}+(1+4x)^{-1}[1.2em]&=1-5x+left(dfrac{-1times(-2)}{1times2}(5x)^{2}right)+…+1-4x[1.2em]&+left(dfrac{-1times(-2)}{1times2}(4x)^{2}right)+…[1.2em]&=1-5x+left(dfrac{2}{2}times25x^{2}right)+…+1-4x[1.2em]&+left(dfrac{-1times(-2)}{1times2}16x^{2}right)+…[1.2em]&=2-9x+25x^{2}+16x^{2}+…[1.2em]&=2-9x+41x^{2}+…end{aligned}$

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Question 4: Find an approximation for $dfrac{1}{0.93^{3}}$ by first expanding $(1-7x)^{-3}$ to five terms.

[8 marks]

$begin{aligned}&(1-7x)^{-3}=1-left(7times(-3)xright)+left(dfrac{-3times(-4)}{1times2}(-7x)^{2}right)[1.2em]&+left(dfrac{-3times(-4)times(-5)}{1times2times3}(-7x)^{3}right)[1.2em]&+left(dfrac{-3times(-4)times(-5)times(-6)}{1times2times3times4}(-7x)^{4}right)+…[1.2em]&=1+21x+left(dfrac{12}{2}times49x^{2}right)+left(dfrac{-60}{6}times(-343)x^{3}right)[1.2em]&+left(dfrac{360}{24}times2401x^{4}right)+…[1.2em]&=1+21x+(6times49x^{2})+(10times343x^{3})+(15times2401x^{4})[1.2em]&+…[1.2em]&=1+21x+294x^{2}+3430x^{3}+36015x^{4}+…end{aligned}$

Use $x=0.01$

$(1-7times0.01)^{-3}=1+(21times0.01)+(294times0.01^{2})+(3430times0.01^{3})+(36015times0.01^{4})+…$

$(1-0.07)^{-3}=1+0.21+(294times0.0001)+(3430times0.000001)+(36015times0.00000001)+…$

$0.93^{-3}=1+0.21+0.0294+0.00343+0.00036015+…$

$begin{aligned}dfrac{1}{0.93^{3}}&=1+0.21+0.0294+0.00343+0.00036015+…[1.2em]&=1.24319015end{aligned}$

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Additional Resources

Exam Tips Cheat Sheet

A Level

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Formula Booklet

A Level

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Specification Points Covered

D1 – Understand and use the binomial expansion of $(a+bx)^n$ for positive integer $n$ ; the notations $n!$ and $nCr$ ; link to binomial probabilities
Extend to any rational $n$ , including its use for approximation; be aware that the expansion is valid for $left|dfrac{bx}{a}right|<1$ . (proof not required)
D6 – Use sequences and series in modelling