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?

hint
For $ x $ such that $ |x|<1$, you might know that
$$F(x)=1+x+x^2+x^3+...=$$
$$\sum_{n=1}^\infty x^n=\frac{1}{1-x}$$
$$xF'(x)=x+2x^2+3x^3+...=$$
$$\sum_{n=1}^\infty nx^n=x\frac{1}{(1-x)^2}$$
differentiate again to get
$$x\frac{ d}{dx}(xF'(x))=$$
$$x+2^2x^2+3^2x^3+...=\sum_{n=1}^{\infty}n^2x^n$$
Now, observe that
$$\frac{(n+1)(n+2)}{2^n}=$$
$$n^2\frac{1}{2^n}+3n\frac{1}{2^n}+2\frac{1}{2^n}=$$
$$n^2x^n+3nx^n+2x^n$$
