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

Question

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.

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

Answer 1

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 $x=\frac{1}{4}$