Binomial Expansion

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 AQA Edexcel OCR

Pascal’s Triangle

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

The numbers on the nth 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 LevelAQAEdexcelOCR

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 LevelAQAEdexcelOCR

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 mth 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 LevelAQAEdexcelOCR
A Level AQA Edexcel OCR

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 LevelAQAEdexcelOCR

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 LevelAQAEdexcelOCR

Example Questions

1s 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

Additional Resources

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