<s>[INST] Evaluation of the series $\sum_{n=1}^{\infty} \frac{n^2+2n+3}{2^n}$. Find the following sum

$$\sum_{n=1}^{\infty} \frac{n^2+2n+3}{2^n}$$

Could someone give some hint to proceed in this question?
 [/INST] Consider that $$ \frac{(n+2)^2}{2^n} - \frac{(n+3)^2}{2^{n+1}} = \frac{n^2+2n-1}{2^n}  = \frac{n^2+2n+3}{2^n} - \frac{4}{2^n}.$$
So, if the required sum is denoted by $S$, we have
$$ \sum_{n=1}^{\infty} \frac{(n+2)^2}{2^n} - \frac{(n+3)^2}{2^{n+1}} = S - 4 \sum_{n=1}^{\infty} \frac{1}{2^n}.$$
The left hand side is a telescoping sum, and on the right hand side we have an infinite geometric series summation.
So we easily get $\dfrac{9}{2} = S - 4$ so that $S = \dfrac{17}{2}$
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