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

Question

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

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How can I use generating functions to solve this?

Answer 1

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 $\color{red}{S=\frac{4}{27}}$.

Answer 2

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

$$\sum_{n=2}^\infty \binom{n}{2} z^n = \frac{\frac14^2}{(1-\frac14)^3} = \frac{4}{27}$$