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 $$\underline{-\dfrac{125}{4}}$$.

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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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