Finding sum of series $\sum^{\infty}_{k=0}\frac{(k+1)(k+2)}{2^k}$ 
Finding sum of series $\displaystyle \sum^{\infty}_{k=0}\frac{(k+1)(k+2)}{2^k}$

what i try
Let $\displaystyle x=\frac{1}{2},$ Then series sum is $\displaystyle \sum^{\infty}_{k=0}(k+1)(k+2)x^k$
$\displaystyle \sum^{\infty}_{k=0}x^k=\frac{1}{1-x}\Rightarrow \sum^{\infty}_{k=0}x^{k+1}=\frac{x}{1-x}=\frac{1-(1-x)}{1-x}$
differentiating both side
$\displaystyle \sum^{\infty}_{k=0}(k+1)x^{k}=+\frac{1}{(1-x)^2}$
How do i solve it help me please

You almost have it. Instead of multiplying the infinite sum by $x$, try multiplying by $x^2$, so you get
$$\sum^{\infty}_{k=0}x^{k+2} = \frac{x^2}{1-x} \tag{1}\label{eq1A}$$
After differentiating twice, you will then get your series sum on the left, i.e., $\sum^{\infty}_{k=0}(k + 1)(k + 2)x^{k}$. You can determine the right side at $x = \frac{1}{2}$. I'll leave the details to you to finish.
