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

Another way:
Let $f(m)=\dfrac{a+bm+cm^2}{2^m}$
and  $\dfrac{(k+1)(k+2)}{2^k}=f(k)-f(k+1)$
so that $$\sum_{k=0}^n(f(k)-f(k+1))=f(0)-f(n+1)$$
$$\implies2(k+1)(k+2)=2(a+bk+ck^2)-[a+b(k+1)+c(k+1)^2]$$
$$\iff4+6k+2k^2=k^2(2c-c)+k(2b-b-2c)+2a-a-b-c$$
Compare the constants and the coefficients of $k,k^2$ to find $a,b,c$
Now we can prove $$\lim_{n\to\infty}f(n+1)=0$$
