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

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. Use what you know about generating functions to determine that value.

How would I do this? My mind is blank. All solutions are highly appreciated!

• Start with the identity $$\sum_{n=0}^\infty x^n=(1-x)^{-1}.$$ Differentiate twice, then multiply by $x^2/2,$ then substitute $x=1/4.$ – bof Oct 25 '16 at 22:20
• Alternatively use the Binomial Theorem: $$(1-x)^{-3}=\sum_{n=0}^\infty(-1)^n\binom{-3}nx^n=\sum_{n=0}^\infty\binom{n+2}2x^n=\sum_{n=2}^\infty\binom n2x^{n-2}=x^{-2}\sum_{n=2}^\infty\binom n2x^n.$$ – bof Oct 25 '16 at 22:49

Hint: $$\binom n2x^n=\frac{n(n-1)x^n}{2!}$$ $${\frac {2x^2}{(1-x)^{3}}}=\sum _{n=2}^{\infty }(n-1)nx^{n}\quad {\text{ for }}|x|<1$$ use $x=\frac{1}{4}$