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?
 A: 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 $S = \dfrac{17}{2}$
A: First, break your problem into smaller sums:
$$\sum_{n = 1}^\infty \frac{n^2}{2^n}+\sum_{n = 1}^\infty \frac{2n}{2^n}+\sum_{n = 1}^\infty\frac{3}{2^n}$$
The third one is just a geometric sum.
You should be able to do this.
Note that we have that:
$$\sum_i q^i = \frac{1}{1-q}$$
We can extend this definition via:
$$\frac{d}{dq}\sum_i q^i = \frac{d}{dq}\frac{1}{1-q}$$
Assuming we can switch the sum with the derivative, we get that:
$$\sum_i iq^{i-1} = \frac{1}{(1-q)^2}$$
To extend this further, we can't just take another derivative (we'd get $\sum_i i(i-1)q^{i-2}$, which isn't what you want exactly).
Instead, we can multipy through by $q$ first, then take the derivative:
$$\sum_i iq^i = \frac{q}{(1-q)^2}$$
Now, we can take the derivative again, to get:
$$\sum_i i^2q^i = \frac{(1-q)^2-q2(1-q)}{(1-q)^4}$$
With these formula, you should be able to compute each individual term (just set $q = \frac{1}{2}$ to get the form you need).
A: HINT:
Let $f(x)$ be given by
$$f(x)=\sum_{n=1}^\infty x^n=\frac{x}{1-x} \tag 1$$
for $|x|<1$.  Then, we have
$$f'(x)=\frac{1}{(1-x)^2} \tag 2$$
and 
$$f''(x)=-\frac{2}{(1-x)^3} \tag 3$$
Using $(1)-(3)$ we see that 
$$\sum_{n=1}^\infty nx^{n}=xf'(x)=\frac{x}{(1-x)^2} \tag 4$$
and 
$$\sum_{n=1}^\infty n^2x^n=x(xf'(x))'=x^2f''(x)+xf'(x)=-\frac{x(1+x)}{(1-x)^3}\tag 5$$
Use the relationships in $(1)$, $(4)$, and $(5)$ with $x=1/2$. 
A: 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 $S=6+7=\color{red}{13}$.
