More Binomial Expansions

More Binomial Expansions

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

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.

A Level AQA Edexcel OCR

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}x\right)^{-1}\\[1.2em]&=-\left(1-(-x)+\dfrac{-1\times(-2)}{1\times2}(-x)^{2}+...\right)\\[1.2em]&+\dfrac{1}{2}\left(1-\dfrac{1}{2}x+\dfrac{-1\times(-2)}{1\times2}\left(\dfrac{1}{2}x\right)^{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}\times\dfrac{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}

A LevelAQAEdexcelOCR

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}\times3x\right)+\left(\dfrac{\dfrac{1}{4}\times -\dfrac{3}{4}}{1\times2}\times(-3x)^{2}\right)\\[1.2em]&+\left(\dfrac{\dfrac{1}{4}\times -\dfrac{3}{4}\times-\dfrac{7}{4}}{1\times2\times3}\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{3\times9}{16\times2}x^{2}-\dfrac{21\times27}{64\times6}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(1-3\times\dfrac{1}{100}\right)^{\frac{1}{4}}=1-\left(\dfrac{3}{4}\times\dfrac{1}{100}\right)-\left(\dfrac{27}{32}\times\left(\dfrac{1}{100}\right)^{2}\right)-\left(\dfrac{189}{128}\times\left(\dfrac{1}{100}\right)^{3}\right)

 

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

A LevelAQAEdexcelOCR

Example Questions

\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}+3\times2^{-1}\left(1+\dfrac{1}{2}x\right)^{-1}\\[1.2em]&+3^{-1}\left(1+\dfrac{1}{3}x\right)^{-1}\\[1.2em]&=2(1+x)^{-1}+3\times\dfrac{1}{2}\left(1+\dfrac{1}{2}x\right)^{-1}\\[1.2em]&+\dfrac{1}{3}\left(1+\dfrac{1}{3}x\right)^{-1}\\[1.2em]&=2(1+x)^{-1}+\dfrac{3}{2}\left(1+\dfrac{1}{2}x\right)^{-1}+\dfrac{1}{3}\left(1+\dfrac{1}{3}x\right)^{-1}\\[1.2em]&=2\left(1-x+\dfrac{-1\times(-2)}{1\times2}x^{2}\right)\\[1.2em]&+\dfrac{3}{2}\left(1-\dfrac{1}{2}x+\dfrac{-1\times(-2)}{1\times2}\left(\dfrac{1}{2}x\right)^{2}\right)\\[1.2em]&+\dfrac{1}{3}\left(1-\dfrac{1}{3}x+\dfrac{-1\times(-2)}{1\times2}\left(\dfrac{1}{3}x\right)^{2}\right)\\[1.2em]&=2\left(1-x+\dfrac{2}{2}x^{2}\right)+\dfrac{3}{2}\left(1-\dfrac{1}{2}x+\dfrac{2}{2}\times\dfrac{1}{4}x^{2}\right)\\[1.2em]&+\dfrac{1}{3}\left(1-\dfrac{1}{3}x+\dfrac{2}{2}\times\dfrac{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}

Try x=0.02

 

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

 

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

 

(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]&=0.9438\end{aligned}

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{-1\times(-2)}{1\times2}(5x)^{2}\right)+...+1-4x\\[1.2em]&+\left(\dfrac{-1\times(-2)}{1\times2}(4x)^{2}\right)+...\\[1.2em]&=1-5x+\left(\dfrac{2}{2}\times25x^{2}\right)+...+1-4x\\[1.2em]&+\left(\dfrac{-1\times(-2)}{1\times2}16x^{2}\right)+...\\[1.2em]&=2-9x+25x^{2}+16x^{2}+...\\[1.2em]&=2-9x+41x^{2}+...\end{aligned}
\begin{aligned}&(1-7x)^{-3}=1-\left(7\times(-3)x\right)+\left(\dfrac{-3\times(-4)}{1\times2}(-7x)^{2}\right)\\[1.2em]&+\left(\dfrac{-3\times(-4)\times(-5)}{1\times2\times3}(-7x)^{3}\right)\\[1.2em]&+\left(\dfrac{-3\times(-4)\times(-5)\times(-6)}{1\times2\times3\times4}(-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+(6\times49x^{2})+(10\times343x^{3})+(15\times2401x^{4})\\[1.2em]&+...\\[1.2em]&=1+21x+294x^{2}+3430x^{3}+36015x^{4}+...\end{aligned}

 

Use x=0.01

 

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

 

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

 

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]&=1.24319015\end{aligned}

Related Topics

MME

Partial Fractions

A Level
MME

Binomial Expansion

A Level
MME

Infinite Series Binomial Expansions

A Level

Additional Resources

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Exam Tips Cheat Sheet

A Level
MME

Formula Booklet

A Level

Worksheet and Example Questions

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