Finding generating function for a given sequence The task is to find the generating function for the sequence $$d_n = (\sum_{k=0}^n \binom{n}{k})*(\sum_{k=0}^n \frac{(k-1)^2}{2^k})$$
let's call this function $D(x)$.
I marked $a_n = \frac{1}{2^n}$; $b_n = \frac{-2n}{2^n}$; $c_n = \frac{n^2}{2^n}$ with respective generating functions $A(x)$, $B(x)$, $C(x)$.
So I found $$A(x) = \frac{1}{1-\frac{x}{2}}$$ and
$$B(x) = \frac{-x}{(1-\frac{x}{2})^2}$$
$$C(x) = \frac{2x+3x^2}{4(1-\frac{x}{2})^3}$$
using derivatives.
now the generating function for $a_n + b_n +c_n$ should be $A(x)+B(x)+C(x)$ and it's $$\frac{x^2}{(1-\frac{x}{2})^3}$$.
Next by using convolution (with let's say $f_n = 1$) we'll get that the function for the series $$u_n = \sum_{k=0}^n \frac{(k-1)^2}{2^k})$$ is $$\frac{x^2}{(1-\frac{x}{2})^3(1-x)}$$
Now since $h_n = \sum_{k=0}^n \binom{n}{k} = 2^n$ the appropriate function is $$H(x) = \frac{1}{1-2x}$$.
Again with using convolution with $h_n$ and $u_n$ we'll get that $$D(x) = H(x)U(x) = \frac{x^2}{(1-\frac{x}{2})^3(1-x)(1-2x)}$$
Is my answer good? or shall i keep expanding it to partial fractions.

Hint: The situation is   somewhat different. 

Write
  \begin{align*}
d_n &= \sum_{k=0}^n \binom{n}{k}\sum_{k=0}^n \frac{(k-1)^2}{2^k}\\
&=\sum_{k=0}^n(k-1)^22^{n-k}
\end{align*}
  and  consider the  Cauchy-product $\sum_{k=0}^na_kb_{n-k}$   when multiplying  generating  functions.

