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!
On
Another approach is to use an arithmetico-geometric sequence, although this isn't using generating functions. To use generating functions, see the accepted answer.
Call the sum $S$, so $$S = \frac{1}{16}+\frac{3}{64}+\frac{6}{256}+\frac{10}{1024}+\dots$$
Divide by $4$, so $$\frac{1}{4}S = \quad \frac{1}{64}+\frac{3}{256}+\frac{6}{1024}+\dots$$
Subtracting the second equation from the first gets $$\frac{3}{4}S = \frac{1}{16}+\frac{2}{64}+\frac{3}{256}+\frac{4}{1024}+\dots$$
We can use a similar process. $$\frac{3}{16}S = \frac{1}{64}+\frac{2}{256}+\frac{3}{1024}+\dots$$
Subtracting the second equation from the first gets $$\frac{9}{16}S = \frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots$$
The RHS is just an infinite geometric sequence. It equals $$\frac{\frac{1}{16}}{1-\frac{1}{4}} = \frac{1}{12}.$$
Now we have $$\frac{9}{16}S = \frac{1}{12},$$
and solving for $S$ gets $$S = \frac{1}{12}\cdot \frac{16}{9} = \boxed{\frac{4}{27}.}$$
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 $x=\frac{1}{4}$