# Infinite Series Binomial Expansions

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## Infinite Series Binomial Expansions

For $(a+bx)^{n}$, we can still get an expansion if $n$ is not a positive whole number. However, the expansion goes on forever. In this page you will find out how to calculate the expansion and how to use it.

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

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## The Infinite Binomial Expansion

Consider writing the binomial coefficients in a different way.

$\begin{pmatrix}n\\0\end{pmatrix}=1$

$\begin{pmatrix}n\\1\end{pmatrix}=\dfrac{n}{1}$

$\begin{pmatrix}n\\2\end{pmatrix}=\dfrac{n(n-1)}{1\times2}$

$\begin{pmatrix}n\\3\end{pmatrix}=\dfrac{n(n-1)(n-2)}{1\times2\times3}$

A clear pattern has emerged. Indeed:

$\begin{pmatrix}n\\r\end{pmatrix}=\dfrac{n(n-1)(n-2)...(n-r+1)}{1\times2\times3\times...\times r}$

We can use this pattern instead of actual binomial coefficients to write an infinite expansion for $(1+ax)^{n}$ when $n$ is not a positive whole number.

$(1+ax)^{n}=1+nax+\dfrac{n(n-1)}{1\times2}a^{2}x^{2}+...\dfrac{n(n-1)...(n-r+1)}{1\times2\times3\times...\times r}a^{r}x^{r}+...$

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@mmerevise

## Note: Factorisation

The formula above only works for expressions of the form $(1+ax)^{n}$, so how do we expand $(ax+b)^{n}$?

$(ax+b)^{n}=b^{n}\left(1+\dfrac{a}{b}x\right)^{n}$, and we can use our formula on $\left(1+\dfrac{a}{b}x\right)^{n}$ then multiply by $b^{n}$.

## Validity of the Binomial Expansion

$(a+bx)^{n}$ is never infinite in value, but an infinite expansion might be unless each term is smaller than the last. To prevent this explosion to infinity we can only work with certain values of $x$. Specifically:

The binomial expansion of $(ax+b)^{n}$ is only valid for $|x|<\left|\dfrac{b}{a}\right|$

Note: If $n$ is an integer then we do not need to worry about this; we get a finite number of terms in the binomial expansion, so it can never have an infinite value and is thus always valid.

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## Combinations of Expansions

Sometimes, we will be asked to expand something that contains more than one binomial term. We do this by treating each binomial term individually, then handling them together at the end.

Example: Find the first three terms of the expansion of $\color{red}(1+2x)^{-1}\color{blue}(1-3x)^{\frac{1}{2}}$

\begin{aligned}\color{red}(1+2x)^{-1}&\color{red}=1-2x+\dfrac{(-1)(-2)}{1\times2}2^{2}x^{2}+...\\[1.2em]&\color{red}=1-2x+4x^{2}+...\end{aligned}

\begin{aligned}\color{blue}(1-3x)^{\frac{1}{2}}&\color{blue}=1+\dfrac{1}{2}(-3)x+\dfrac{\dfrac{1}{2}\times\dfrac{-1}{2}}{1\times2}(-3)^{2}x^{2}+...\\[1.2em]&\color{blue}=1-\dfrac{3}{2}x-\dfrac{\dfrac{1}{4}}{2}\times9x^{2}+...\\[1.2em]&\color{blue}=1-\dfrac{3}{2}x-\dfrac{9}{8}x^{2}+...\end{aligned}

\begin{aligned}&\color{red}(1+2x)^{-1}\color{blue}(1-3x)^{\frac{1}{2}}\color{grey}=\\[1.2em]&\color{red}(1-2x+4x^{2}+...)\color{blue}\left(1-\dfrac{3}{2}x-\dfrac{9}{8}x^{2}+...\right)\\[1.2em]&\color{grey}=\color{red}1\color{blue}\left(1-\dfrac{3}{2}x-\dfrac{9}{8}x^{2}+...\right)\color{grey}-\color{red}2x\color{blue}\left(1-\dfrac{3}{2}x+...\right)\color{grey}\\[1.2em]&+\color{red}4x^{2}\color{blue}(1+...)\\[1.2em]&\color{grey}=1-\dfrac{3}{2}x-\dfrac{9}{8}x^{2}-2x+3x^{2}+4x^{2}+...\\[1.2em]&\color{grey}=1-\dfrac{7}{2}x+\dfrac{47}{8}x^{2}+...\end{aligned}

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## Infinite Series Binomial Expansions Example Questions

i) $(3+5x)^{3}=3^{3}\left(1+\dfrac{5}{3}x\right)^{3}=27\left(1+\dfrac{5}{3}x\right)^{3}$

ii) $(2+x)^{6}=2^{6}\left(1+\dfrac{1}{2}x\right)^{6}=64\left(1+\dfrac{1}{2}x\right)^{6}$

iii) $(81+18x)^{\frac{1}{4}}=81^{\frac{1}{4}}\left(1+\dfrac{18}{81}\right)^{\frac{1}{4}}=3\left(1+\dfrac{2}{9}\right)^{\frac{1}{4}}$

iv) $(3+2x)^{\frac{1}{2}}=3^{\frac{-1}{2}}\left(1+\dfrac{2}{3}x\right)^{\frac{-1}{2}}=\dfrac{1}{\sqrt{3}}\left(1+\dfrac{2}{3}x\right)^\frac{-1}{2}$

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\begin{aligned}(1+4x)^{-2}&=1+\big( (-2)\times4x\big) +\left( \dfrac{-2\times(-3)}{1\times2}(4x)^{2}\right) +\left( \dfrac{-2\times(-3)\times(-4)}{1\times2\times3}(4x)^{3}\right) \\[1.2em]&=1-8x+\left( \dfrac{6}{2}\times4^{2}x^{2}\right) -\left( \dfrac{24}{6}\times4^{3}x^{3}\right) \\[1.2em]&=1-8x+\big( 3\times16x^{2}\big) -\big( 4\times64x^{3}\big) \\[1.2em]&=1-8x+48x^{2}-256x^{3}\end{aligned}

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\begin{aligned}(3+x)^{\frac{1}{2}}&=3^{\frac{1}{2}}\left(1+\dfrac{1}{3}x\right)^{\frac{1}{2}}\\[1.2em]&=\sqrt{3}\left(1+\left( \dfrac{1}{2}\times\dfrac{1}{3}x\right) +\left( \dfrac{\dfrac{1}{2}\times\dfrac{-1}{2}}{1\times2}\times\left(\dfrac{1}{3}x\right)^{2}\right) \right) \\[1.2em]&=\sqrt{3}\left(1+\dfrac{1}{6}x-\left( \dfrac{\left( \dfrac{1}{4}\right) }{2}\times\dfrac{1}{9}x^{2}\right) \right) \\[1.2em]&=\sqrt{3}\left(1+\dfrac{1}{6}x-\left( \dfrac{1}{8}\times\dfrac{1}{9}x^{2}\right) \right) \\[1.2em]&=\sqrt{3}\left(1+\dfrac{1}{6}x-\dfrac{1}{72}x^{2}\right)\end{aligned}

This is valid for $|x|<3$

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\begin{aligned}&\dfrac{(1+2x)^{\frac{1}{2}}}{(3+2x)^{2}}=(1+2x)^{\frac{1}{2}}(3+2x)^{-2}\\[1.2em]&=\left(1+\left( \dfrac{1}{2}\times2x\right) +\left( \dfrac{\dfrac{1}{2}\times\dfrac{-1}{2}}{1\times2}(2x)^{2}\right) +...\right)\times3^{-2}\left(1+\dfrac{2}{3}x\right)^{-2}\\[1.2em]&=\left(1+x-\left( \dfrac{\left( \dfrac{1}{4}\right) }{2}\times4x^{2}\right) +...\right)\times\dfrac{1}{9}\left(1-\left( 2\times\dfrac{2}{3}x\right) +\left( \dfrac{-2\times(-3)}{1\times2}\left(\dfrac{2}{3}x\right)^{2}\right) +...\right)\\[1.2em]&=\dfrac{1}{9}\left(1+x-\left( \dfrac{1}{8}\times4x^{2}\right) +...\right)\left(1-\dfrac{4}{3}x+\dfrac{6}{2}\times\dfrac{4}{9}x^{2}+...\right)\\[1.2em]&=\dfrac{1}{9}\left(1+x-\dfrac{1}{2}x^{2}+...\right)\left(1-\dfrac{4}{3}x+\dfrac{4}{3}x^{2}+...\right)\\[1.2em]&=\dfrac{1}{9}\left(1+x-\dfrac{1}{2}x^{2}-\dfrac{4}{3}x-\dfrac{4}{3}x^{2}+\dfrac{4}{3}x^{2}+...\right)\\[1.2em]&=\dfrac{1}{9}\left(1-\dfrac{1}{3}x-\dfrac{1}{2}x^{2}+...\right)\\[1.2em]&=\dfrac{1}{9}-\dfrac{1}{27}x-\dfrac{1}{18}x^{2}+...\end{aligned}

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