# Binomial Expansion

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

## Binomial Expansion

Binomial expansion uses binomial coefficients to expand two terms in brackets of the form $(ax+b)^{n}$. When $n$ is a positive whole number the expansion is finite. When $n$ is not, the expansion is infinite.

A Level   ## Pascal’s Triangle Pascal’s triangle is made by putting $1$s down the sides, then every number comes from adding together the two above it.

The numbers on the $n$th row are the coefficients of $(1+x)^{n}$

$(1+x)^{0}=1$

$(1+x)^{1}=1+x$

$(1+x)^{2}=1+2x+x^{2}$

$(1+x)^{3}=1+3x+3x^{2}+x^{3}$

$(1+x)^{4}=1+4x+6x^{2}+4x^{3}+x^{4}$

and so on.

A Level   ## Binomial Coefficients The binomial coefficient $\begin{pmatrix}n\\m\end{pmatrix}=\dfrac{n!}{m!(n-m)!}$ where $m$ and $n$ are whole numbers and $n>m$. We can also write them as $\text{}^{n}C_{m}$.

The binomial coefficients are the same as the numbers in Pascal’s triangle.

This means we can write our expansion as:

$(1+x)^{n}=\begin{pmatrix}n\\0\end{pmatrix}+\begin{pmatrix}n\\1\end{pmatrix}x+\begin{pmatrix}n\\2\end{pmatrix}x^{2}+...+\begin{pmatrix}n\\n-1\end{pmatrix}x^{n-1}+\begin{pmatrix}n\\n\end{pmatrix}x^{n}$

A Level   ## General Binomial Expansion Formula

So far we have only seen how to expand $(1+x)^{n}$, but ideally we want a way to expand more general things, of the form $(a+b)^{n}$. In this expansion, the $m$th term has powers $a^{m}b^{n-m}$. We can use this, along with what we know about binomial coefficients, to give the general binomial expansion formula.

$(a+b)^{n}=\begin{pmatrix}n\\0\end{pmatrix}a^{n}+\begin{pmatrix}n\\1\end{pmatrix}a^{n-1}b+\begin{pmatrix}n\\2\end{pmatrix}a^{n-2}b^{2}+...+\begin{pmatrix}n\\m\end{pmatrix}a^{m}b^{n-m}+...+\begin{pmatrix}n\\n-1\end{pmatrix}ab^{n-1}+\begin{pmatrix}n\\n\end{pmatrix}b^{n}$

A Level   A Level   ## Example 1: Binomial Expansion

Expand $(1+2x)^{2}$

[2 marks]

$(1+2x)^{2}=\begin{pmatrix}2\\0\end{pmatrix}1^{2}+\begin{pmatrix}2\\1\end{pmatrix}1\times2x+\begin{pmatrix}2\\2\end{pmatrix}(2x)^{2}$

Find the binomial coefficients:

$\begin{pmatrix}2\\0\end{pmatrix}=1$

$\begin{pmatrix}2\\1\end{pmatrix}=2$

$\begin{pmatrix}2\\2\end{pmatrix}=1$

Substitute binomial coefficients into equation:

\begin{aligned}(1+2x)^{2}&=1\times1^{2}+2\times1\times2x+1\times(2x)^{2}\\[1.2em]&=1\times1+2\times2x+1\times4x^{2}\\[1.2em]&=1+4x+4x^{2}\end{aligned}

A Level   ## Example 2: Finding Coefficients

Find the coefficient of $x^{3}$ in $(x+8)^{6}$.

[2 marks]

$x^{3}$ term is $\begin{pmatrix}6\\3\end{pmatrix}x^{3}8^{6-3}$

$\begin{pmatrix}6\\3\end{pmatrix}=20$

$8^{6-3}=8^{3}=512$

$x^{3}$ term is $512\times20x^{3}$

$x^{3}$ term is $10240x^{3}$

Coefficient of $x^{3}$ is $10240$

A Level   ## Example Questions

$1$s down the sides.

$1+6=7$ so $7$ is the next number in.

$6+15=21$ so $21$ is the next number in.

$15+20=35$ so $35$ is the next number in.

So the whole row is:

$1,7,21,35,35,21,7,1$
$\left(\dfrac{1}{4}-2x\right)^{4}=\begin{pmatrix}4\\0\end{pmatrix}(-2x)^{4}+\begin{pmatrix}4\\1\end{pmatrix}\dfrac{1}{4}(-2x)^{3}+\begin{pmatrix}4\\2\end{pmatrix}\left(\dfrac{1}{4}\right)^{2}(-2x)^{2}+\begin{pmatrix}4\\3\end{pmatrix}\left(\dfrac{1}{4}\right)^{3}(-2x)+\begin{pmatrix}4\\4\end{pmatrix}\left(\dfrac{1}{4}\right)^{4}$

$\begin{pmatrix}4\\0\end{pmatrix}=1$

$\begin{pmatrix}4\\1\end{pmatrix}=4$

$\begin{pmatrix}4\\2\end{pmatrix}=6$

$\begin{pmatrix}4\\3\end{pmatrix}=4$

$\begin{pmatrix}4\\1\end{pmatrix}=1$

$\left(\dfrac{1}{4}-2x\right)^{4}=\big( 1\times(-2x)^{4}\big) +\big( 4\times\dfrac{1}{4}(-2x)^{3}\big) +\big( 6\times\left(\dfrac{1}{4}\right)^{2}(-2x)^{2}\big) +\big( 4\times\left(\dfrac{1}{4}\right)^{3}(-2x)\big) +\big( 1\times\left(\dfrac{1}{4}\right)^{4}\big)$

$\left(\dfrac{1}{4}-2x\right)^{4}=\big( 1\times16x^{4}\big) +\big( 4\times\dfrac{1}{4}\times(-8)x^{3}\big) +\big( 6\times\dfrac{1}{16}\times4x^{2}\big) +\big( 4\times\dfrac{1}{64}\times(-2)x\big) +\big( 1\times\dfrac{1}{256}\big)$

$\left(\dfrac{1}{4}-2x\right)^{4}=16x^{4}-8x^{3}+\dfrac{24}{16}x^{2}-\dfrac{8}{64}x+\dfrac{1}{256}$

$\left(\dfrac{1}{4}-2x\right)^{4}=16x^{4}-8x^{3}+\dfrac{3}{2}x^{2}-\dfrac{1}{8}x+\dfrac{1}{256}$

$x^{4}$ term is $\begin{pmatrix}9\\4\end{pmatrix}(5x)^{4}(-2)^{5}=-\begin{pmatrix}9\\4\end{pmatrix}5^{4}x^{4}2^{5}$

$\begin{pmatrix}9\\4\end{pmatrix}=126$

$5^{4}=625$

$2^{5}=32$

$x^{3}$ coefficient is $-126\times625\times32=-2520000$

$x^{2}$ coefficient is $\begin{pmatrix}3\\2\end{pmatrix}\left(\dfrac{2}{3}\right)^{2}p$

$\begin{pmatrix}3\\2\end{pmatrix}=3$

$\left(\dfrac{2}{3}\right)^{2}=\dfrac{4}{9}$

Hence, $3\times\dfrac{4}{9}\times p=4$

$\dfrac{4}{3}p=4$

$p=4\times\dfrac{3}{4}$

$p=3$

A Level

A Level

A Level

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