Sum with Generating Functions Find the sum
$$\sum_{n=2}^{\infty} \frac{\binom n2}{4^n} ~~=~~ \frac{\binom 22}{16}+\frac{\binom 32}{64}+\frac{\binom 42}{256}+\cdots$$

How can I use generating functions to solve this?
 A: If $f(z) = \sum\limits_{n=0}^\infty a_n z^n$ converges for $|z| < \rho$, then for any $m \ge 0$,
$$\frac{z^m}{m!} \frac{d^m}{dz^m} f(z) = \sum_{n=0}^\infty a_m\binom{n}{m} z^n
\quad\text{ for } |z| < \rho.$$
Apply this to $$f(z) = \frac{1}{1-z} = \sum_{n=0}^\infty z^n,$$ we get
$$\sum_{n=2}^\infty \binom{n}{2} z^n = \frac{z^2}{2!}\frac{d^2}{dz^2}\frac{1}{1-z} = \frac{z^2}{(1-z)^3}$$
Substitute $z = \frac14$ will give you
$$\sum_{n=2}^\infty \binom{n}{2} z^n = \frac{\frac14^2}{(1-\frac14)^3} = \frac{4}{27}$$
A: For an alternative technique, let
$$ S = \sum_{n\geq 2}\frac{\binom{n}{2}}{4^n}.$$
We have:
$$\begin{eqnarray*} 3S = 4S-S &=& \sum_{n\geq 2}\frac{\binom{n}{2}}{4^{n-1}}-\sum_{n\geq 2}\frac{\binom{n}{2}}{4^{n}}=\frac{1}{4}+\sum_{n\geq 2}\frac{\binom{n+1}{2}-\binom{n}{2}}{4^n}\\&=&\frac{1}{4}+\sum_{n\geq 2}\frac{n}{4^n}.\end{eqnarray*}$$
If we set:
$$ T = \sum_{n\geq 2}\frac{n}{4^n}$$
we have:
$$ \begin{eqnarray*}3T = 4T-T = \sum_{n\geq 2}\frac{n}{4^{n-1}}-\sum_{n\geq 2}\frac{n}{4^n}=\frac{1}{2}+\sum_{n\geq 2}\frac{1}{4^n}=\frac{7}{12}\end{eqnarray*} $$
hence $T=\frac{7}{36}$ and $3S=\frac{4}{9}$, so $\color{red}{S=\frac{4}{27}}$.
