<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] The given series is absolutely convergent, and by setting $S=\sum_{n\geq 1}\frac{n^2+2n+3}{2^n}$ we also get:
$$ 2S = \sum_{n\geq 1}\frac{n^2+2n+3}{2^{n-1}} = 6+\sum_{n\geq 1}\frac{n^2+4n+6}{2^n}\tag{1}$$
from which:
$$ S = 2S-S = 6+\sum_{n\geq 1}\frac{2n+3}{2^n}\tag{2} $$
and by setting $T=\sum_{n\geq 1}\frac{2n+3}{2^n}$ we also get:
$$ 2T = \sum_{n\geq 1}\frac{2n+3}{2^{n-1}} = 5+\sum_{n\geq 1}\frac{2n+5}{2^n}\tag{3} $$
$$ T = 2T-T = 5+\sum_{n\geq 1}\frac{2}{2^n} = 7 \tag{4} $$
so $S=6+7=\color{red}{13}$.
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