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Binomial expansion I

Solving binomial problems

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Summary

Solving binomial problems

In a nutshell

It's possible to use parts of the binomial expansion formula to work out the coefficient of a general term in $(x+y)^n$ .

The general term

The general term in the expansion of $(x+y)^n$ is equal to $\binom{n}{r}x^{n-r}y^r$ . This is the case as there are $\binom{n}{r}$ ways of selecting $r$ copies of $y$ from the $n$ bracketed terms.

Example 1

What is the coefficient of $t^3$ in the expansion of $(5-\dfrac{t}{2})^5$ ?

The general term of $(x+y)^n$ is equal to $\binom{n}{r}x^{n-r}y^r$ . Substitute $n = 5$ , $r = 3$ , $x = 5$ and $y = (-\dfrac{t}{2})$ into this expression:

$\begin{aligned}\binom{n}{r}x^{n-r}y^r&= \binom{5}{3}5^{(5-3)}(-\dfrac{t}{2})^3 \\&= 10 \times 25 \times -\dfrac{t^3}{8} \\&=-\dfrac{125}{4}t^3\end{aligned}$ 

Therefore, the coefficient of $t^3$ in the expansion of $\Big(5-\dfrac{t}{2}\Big)^5$ is $\underline{-\dfrac{125}{4}}$ .

Example 2

Given that the coefficient of $t^4$ in the expansion of $(3 - st)^8$ is equal to $\dfrac{315}{2}$ , find two possible values of $s$ .

Substitute $n=8$ , $r=4$ , $x = 3$ and $y= -st$ into the general term to get:

$\begin{aligned}\binom{n}{r}x^{n-r}y^r&=\binom{8}{4}3^4(-st)^4 \\ &= 70 \times 81s^4t^4\\&= 5670s^4t^4\end{aligned}$

Rearrange the expression $\dfrac{315}{2} = 5670s^4$ so that $s^4$ is the subject to get:

$s^4 = \dfrac{1}{36}$

Deduce that $s^2 = \pm \dfrac{1}{6}$ , and so two possible values of $s$ are given by $\pm \dfrac{1}{\sqrt{6}}$ , as these square to $\dfrac{1}{6}$ .

Rationalise the denominator of $\pm\dfrac{1}{\sqrt{6}}$ to get:

$\begin{aligned}\pm\dfrac{1}{\sqrt{6}} &= \pm\dfrac{1 \times \sqrt{6}}{\sqrt{6} \times \sqrt{6}} \\&= \pm\dfrac{\sqrt{6}}{6}\end{aligned}$ 

Therefore, $2$ possible values for $s$ are $\underline{\pm\dfrac{\sqrt{6}}{6}}$ .

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Binomial expansion

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