Finding the coefficient of x^n generating functions

Question

Find the cofficient of $x^n$ in the expansion of $b^m x^m \over (1−bx)^{m+1}$

Here b is a real number. Note your answer may depend on conditions involving m and n.

I started off by isolating $1\over(1-bx)^{m+1}$ to make it fit the generating function $$\sum_{n=0}^\infty b^n {n+m \choose n} x^n $$

But then after adding back in the $b^m$ and the $x^m$ I get stuck here $$\sum_{n=0}^\infty b^n {n+m \choose n} x^n b^m x^m $$

I know from here it should be a matter of simple algebra but I seem to be stuck.

Answer 1

First,

$$\begin{align*} \frac{b^mx^m}{(1-bx)^{m+1}}&=b^mx^m\sum_{n\ge 0}\binom{n+m}nb^nx^n\\ &=\sum_{n\ge 0}\binom{n+m}nb^{m+n}x^{m+n}\;. \end{align*}$$

Now let $k=m+n$; then

$$\sum_{n\ge 0}\binom{n+m}nb^{m+n}x^{m+n}=\sum_{k\ge m}\binom{k}{k-m}b^kx^k\;,$$

and you can rename $k$ back to $n$ to write

$$\frac{b^mx^m}{(1-bx)^{m+1}}=\sum_{n\ge m}\binom{n}{n-m}b^nx^n\;.$$

The coefficient of $x^n$ is therefore $\dbinom{n}{n-m}b^n$. Note that this is correct even for $0\le n<m$, since in those cases the binomial coefficient is $0$.

Answer 2

You are almost right. Your problem is that you used $n$ in your generating function, while $n$ is your input parameter.

To make this clear, I will rewrite your series using $k$:

$\begin{array}\\ \sum_{k=0}^\infty b^k {k+m \choose k} x^k b^m x^m &=\sum_{k=0}^\infty b^{k+m} {k+m \choose k} x^{k+m}\\ &=\sum_{k=m}^\infty b^{k} {k \choose k-m} x^{k}\\ &=\sum_{k=m}^\infty b^{k} {k \choose m} x^{k}\\ \end{array} $

From this you can see that the coefficient of $x^n$ is zero if $n < m$ and $b^{n} {n \choose m} $ if $n \ge m$.