Finding the coefficient of x^n generating functions 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. 
 A: 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 $0$.
A: 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
$n \ge m$.
