Use generating function to find coefficient

Question

Use a generating function to find the coefficient of $x^{22}$ in:

$$\frac{1+3x}{(1-x)^8}$$

I know I need to use a binomial expansion on the lower term, but what about the upper term?

Answer 1

Recall that $$\frac1{1-x}=\sum_{n=0}^\infty x^n.$$ So $$\frac{\mathsf d^7}{\mathsf dx^7}\left[\frac1{1-x}\right] = \frac{7!}{(1-x)^8}$$ and $$\begin{align*} \frac{\mathsf d^7}{\mathsf dx^7}\sum_{n=0}^\infty x^n =\sum_{n=7}^\infty \frac{n!}{(n-7)!}x^{n-7}=\sum_{n=0}^\infty\frac{(n+7)!}{n!}x^{n} \end{align*}$$ Hence $$\begin{align*}\frac{1+3x}{(1-x)^8} &= \frac {1+3x}{8!}\frac{\mathsf d^7}{\mathsf dx^7}\left[\frac1{1-x}\right]\\ &= \frac{1+3x}{7!}\sum_{n=0}^\infty\frac{(n+7)!}{n!}x^n\\ &= \sum_{n=0}^\infty\frac{(n+7)!}{n!7!}x^n + \sum_{n=0}^\infty \frac{3(n+7)!}{n!7!}x^{n+1}\\ &= \sum_{n=0}^\infty\binom {n+7}7 x^n + \sum_{n=1}^\infty 3\binom {n+6}7x^n\\ &= 1 + \sum_{n=1}^\infty \left(\binom{n+7}7+3\binom{n+6}7\right)x^n \end{align*} $$ So the coefficient of $x^{22}$ is $$\binom {22+7}{7} + 3\binom{22+6}7 = 5112900.$$