Does $\sum_{n=0}^{\infty} \frac{(n+1)(n+2)}{2^n}=16$? Here's my question, it is rather straightforward: does $\sum_{n=0}^{\infty} \frac{(n+1)(n+2)}{2^n}=16$? First I tested if $\sum_{n=0}^{\infty} \frac{(n+1)(n+2)}{2^n}$ diverges to make sure it doesn't add to $\infty$. I found that taking the limit of the series as n approaches $\infty$ gives no information.
Then I expanded the series to get $\frac{2}{1}+\frac{6}{2}+\frac{12}{4}+....$ and found no pattern. My goal is to evaluate the series and find out if it's equal to 16 or not. How should I do that? or is there another approach to the question?

$$S:=\sum_{n=0}^{\infty} \frac{(n+1)(n+2)}{2^n}$$
$$\frac S2:=\sum_{n=0}^{\infty} \frac{(n+1)(n+2)}{2^{n+1}}=\sum_{n=0}^{\infty} \frac{n(n+1)}{2^n}$$
so that
$$S-\frac S2=\sum_{n=0}^{\infty} \frac{2(n+1)}{2^n}.$$
Then
$$T:=\sum_{n=0}^{\infty} \frac{n+1}{2^n}$$
and
$$T-\frac T2=\sum_{n=0}^{\infty} \frac1{2^n}.$$
