1.6.1 Binomial Expansion - GCEALevelMaths9709

In this topic we will learn how to:

use the expansion of $\left(a + b\right)^{n}$ , where $n$ is a positive integer

We can expand $\left(a + b\right)^{n}$ using the formula below,
$\left(a + b\right)^{n} = a^{n} + \begin{pmatrix}n \\ 1\end{pmatrix}a^{n - 1}b + \begin{pmatrix}n \\ 2\end{pmatrix}a^{n - 2}b^{2} + \begin{pmatrix}n \\ 3\end{pmatrix}a^{n - 3}b^{3}… + b^{n}$ Where $n$ is a positive integer.

Note: $\begin{pmatrix}n \ r\end{pmatrix}$ represents $^{n}C{r}$ , the function is available on your calculator, possibly as $nCr$ .

Let’s look at some past paper questions on binomial expansion.

1. (a) Expand $\left(1 + a\right)^{5}$ in ascending powers of $a$ up to and including the term in $a^{3}$ . (9709/13/M/J/20 number 4)

Use the formula for binomial expansion,
$\left(a + b\right)^{n} = a^{n} + \begin{pmatrix}n \\ 1\end{pmatrix}a^{n - 1}b + \begin{pmatrix}n \\ 2\end{pmatrix}a^{n - 2}b^{2} + \begin{pmatrix}n \\ 3\end{pmatrix}a^{n - 3}b^{3}... + b^{n}$ $\left(1 + a\right)^{5} = 1^{n} + \begin{pmatrix}5 \\ 1\end{pmatrix}\left(1\right)^{4}\left(a\right) + \begin{pmatrix}5 \\ 2\end{pmatrix}\left(1\right)^{3}\left(a\right)^{2} + \begin{pmatrix}5 \\ 3\end{pmatrix}\left(1\right)^{2}\left(a\right)^{3}$ $\left(1 + a\right)^{5} = 1 + 5a + 10a^{2} + 10a^{3}$ Therefore, the final answer is,
$1 + 5a + 10a^{2} + 10a^{3}$ (b) Hence expand $[1 + \left(x + x^{2}\right)]^{5}$ in ascending powers of $x$ up to and including the term in $x^{3}$ , simplifying your answer.

Notice how this is very similar to the expression above. $a$ has been replaced with $x + x^{2}$ ,
$\left(1 + \textcolor{#2192ff}{a}\right)^{5}$ $[1 + \textcolor{#2192ff}{\left(x + x^{2}\right)}]^{5}$ In the expansion of $\left(1 + a\right)^{5}$ , substitute $a$ with $x + x^{2}$ ,
$1 + 5\textcolor{#2192ff}{a} + 10\textcolor{#2192ff}{a}^{2} + 10\textcolor{#2192ff}{a}^{3}$ $1 + 5\left(\textcolor{#2192ff}{x + x^{2}}\right) + 10\left(\textcolor{#2192ff}{x + x^{2}}\right)^{2} + 10\left(\textcolor{#2192ff}{x + x^{2}}\right)^{3}$ $1 + 5x + 5x^{2} + 10\left(x + x^{2}\right)^{2} + 10\left(x + x^{2}\right)^{3}$ Expand the quadratic and simplify,
$1 + 5x + 5x^{2} + 10\left(x^{2} + 2x^{3} + x^{4}\right) + 10\left(x + x^{2}\right)^{3}$ $1 + 5x + 5x^{2} + 10x^{2} + 20x^{3} + 10x^{4} + 10\left(x + x^{2}\right)^{3}$ $1 + 5x + 15x^{2} + 20x^{3} + 10x^{4} + 10\left(x + x^{2}\right)^{3}$ Disregard the term in $x^{4}$ ,
$1 + 5x + 15x^{2} + 20x^{3} + 10\left(x + x^{2}\right)^{3}$ Expand the cubic, disregarding any terms with a power in $x$ higher than $3$ ,
$1 + 5x + 15x^{2} + 20x^{3} + 10x^{3}$ Note: Only the first term in the cubic has a power in $x$ of $3$ or lower.

Simplify,
$1 + 5x + 15x^{2} + 30x^{3}$ Therefore, the final answer is,
$1 + 5x + 15x^{2} + 30x^{3}$ 2. The coefficient of $x^{3}$ in the expansion of $\left(1 + kx\right)\left(1 - 2x\right)^{5}$ is $20$ . Find the value of the constant $k$ . (9709/12/O/N/20 number 1)
$\left(1 + kx\right)\left(1 - 2x\right)^{5}$ Let’s start by expanding $\left(1 - 2x\right)^{5}$ up to and including the term in $x^{3}$ ,
$\left(1 - 2x\right)^{5}$ $\left(a + b\right)^{n} = a^{n} + \begin{pmatrix}n \\ 1\end{pmatrix}a^{n - 1}b + \begin{pmatrix}n \\ 2\end{pmatrix}a^{n - 2}b^{2} + \begin{pmatrix}n \\ 3\end{pmatrix}a^{n - 3}b^{3}... + b^{n}$ $\left(1 - 2x\right)^{5} = 1^{5} + \begin{pmatrix}5 \\ 1\end{pmatrix}\left(1^{4}\right)\left(-2x\right) + \begin{pmatrix}5 \\ 2\end{pmatrix}\left(1\right)^{3}\left(-2x\right)^{2} + \begin{pmatrix}5 \\ 3\end{pmatrix}\left(1\right)^{2}\left(-2x\right)^{3}$ $\left(1 - 2x\right)^{5} = 1 - 10x + 40x^{2} - 80x^{3}$ Substitute $\left(1 - 2x\right)^{5}$ with the expansion,
$\left(1 + kx\right)\textcolor{#2192ff}{\left(1 - 2x\right)^{5}}$ $\left(1 + kx\right)\textcolor{#2192ff}{\left(1 - 10x + 40x^{2} - 80x^{3}\right)}$ Identify the terms that give a term in $x^{3}$ when multiplied and add them together,
$\left(1 \times \left(-80x^{3}\right)\right) + \left(kx \times 40x^{2}\right)$ $- 80x^{3} + 40kx^{3}$ $\left(-80 + 40k\right)x^{3}$ Note: If you’re struggling to understand this, fully expand the two brackets together and pick out the terms in $x^{3}$ .

Since the coefficient of $x^{3}$ is $20$ , equate $-80 + 40k$ to $20$ ,
$-80 + 40k = 20$ $40k = 80 + 20$ $40k = 100$ $k = \frac{5}{2}$ Therefore, the final answer is,
$k = \frac{5}{2}$ 3. (a) Find the coefficient of $x^{2}$ in the expansion of $\left(x - \frac{2}{x}\right)^{6}$ . ( 9709/12/M/J/20 number 1)
$\left(\textcolor{#2192ff}{x}\ \textcolor{#0f0}{-\ \frac{2}{x}}\right)^{6}$ For us to get a term in $x^{2}$ , $\textcolor{#2192ff}{a}$ has to be to the power $4$ and $\textcolor{#0f0}{b}$ to the power $2$ , so that they cancel out to give a term in $x^{2}$ , hence,
Note: Remember the powers of $a$ and $b$ should always add up to $n$ .
$\begin{pmatrix} n \\ 2 \end{pmatrix} \times \textcolor{#2192ff}{a}^{4} \times \textcolor{#0f0}{b}^{2}$ $\begin{pmatrix} 6 \\ 2 \end{pmatrix} \times \textcolor{#2192ff}{x}^{4} \times \left(\textcolor{#0f0}{-\frac{2}{x}}\right)^{2}$ $15x^{4}\left(\frac{4}{x^{2}}\right)$ $60x^{2}$ Therefore, the coefficient of $x^{2}$ is,
$60$ (b) Find the coefficient of $x^{2}$ in the expansion of $\left(2 + 3x^{2}\right)\left(x - \frac{2}{x}\right)^{6}$ .
$\left(2 + 3x^{2}\right)\left(x - \frac{2}{x}\right)^{6}$ $\left(\textcolor{#2192ff}{x}\ \textcolor{#0f0}{-\ \frac{2}{x}}\right)^{6}$ We have to start by finding the term independent of $x$ in the expansion of $\left(x - \frac{2}{x}\right)^{6}$ . For us to get a term independent of $x$ , the $x$ in $\textcolor{#2192ff}{a}$ and $\textcolor{#0f0}{b}$ should cancel out when they multiply, therefore, $\textcolor{#2192ff}{a}$ should be to the power $3$ and $\textcolor{#0f0}{b}$ should be to the power $3$ ,
$\begin{pmatrix} n \\ 3\end{pmatrix} \times \textcolor{#2192ff}{a}^{3} \times \textcolor{#0f0}{b}^{3}$ $\begin{pmatrix} 6 \\ 3\end{pmatrix} \times \textcolor{#2192ff}{x}^{3} \times \left(\textcolor{#0f0}{-\frac{2}{x}}\right)^{3}$ $20x^{3}\left(-\frac{8}{x^{3}}\right)$ $-160$ Now substitute $\left(x - \frac{2}{x}\right)^{6}$ with the term independent of $x$ and the term in $x^{2}$ ,
$\left(2 + 3x^{2}\right)\textcolor{#2192ff}{\left(x - \frac{2}{x}\right)^{6}}$ $\left(2 + 3x^{2}\right)\textcolor{#2192ff}{\left(60x^{2} - 160\right)}$ Identify the terms that give you a term in $x^{2}$ when multiplied and add them together,
$2\left(60x^{2}\right) + 3x^{2}\left(-160\right)$ $120x^{2} - 480x^{2}$ $-360x^{2}$ Therefore, the coefficient of $x^{2}$ is,
$-360$

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