Generating Function Sum and Combinotorics Problem:
The sum $\sum_{n=2}^{\infty} \frac{\binom n2}{4^n} ~~=~~ \frac{\binom 22}{16}+\frac{\binom 32}{64}+\frac{\binom 42}{256}+\cdots$ has a finite value. Determine that value.

I am quite stuck on how to do this. Can somebody give me <only> a hint or hints to get going? 
Thanks!

Consider
$$S=\sum_{n=2}^\infty \binom{n}{2}x^n =\frac 12 \sum_{n=2}^\infty n(n+1) x^n=\frac 12 \sum_{n=2}^\infty[n(n-1)+2n] x^n$$
$$S=\frac 12 \sum_{n=2}^\infty n(n-1) x^n+ \sum_{n=2}^\infty n x^n=\frac {x^2}2 \sum_{n=2}^\infty n(n-1) x^{n-2}+x\sum_{n=2}^\infty n x^{n-1}$$
$$S=\frac {x^2}2 \left(\sum_{n=2}^\infty  x^n \right)''+x\left(\sum_{n=2}^\infty  x^n \right)'$$ When finished, let $x=\frac 14$.
