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

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.

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 $$\lim_{n\to\infty}f(n+1)=0$$