Solve summation of $\sum_{j=0}^{n-2}2^j (n-j)$ Question
While Solving a recursive equation , i am stuck at this summation and unable to move forward.Summation is 

$$\sum_{j=0}^{n-2}2^j (n-j)$$

My Approach

$$\sum_{j=0}^{n-2}2^j (n-j) = \sum_{j=0}^{n-2}2^j \times n-\sum_{j=0}^{n-2} 
2^{j} \times j$$

$$=n \times (2^{n-1}-1)-\sum_{j=0}^{n-2} 
2^{j} \times j$$
I am unable to move forward , please help me out!

In general:
\begin{align}
\sum_{k=0}^{n} x^{k} &= \frac{1 - x^{n+1}}{1-x} \\
\sum_{k=0}^{n} k \, x^{k} &= \frac{x \, (1 - (n+1) x^{n} + n x^{n+1})}{(1-x)^{2}}
\end{align}
Now
\begin{align}
\sum_{k=0}^{n} k \, \frac{1}{x^{k}} &= \frac{n - (n+1) x + x^{n+1}}{x^{n} \, (1-x)^{2}} \\
\sum_{k=0}^{n} k \, x^{n-k} &= \frac{n - (n+1) x + x^{n+1}}{ (x-1)^{2}}.
\end{align}
Using
$$\sum_{k=0}^{n} k \, x^{n-k} = \sum_{k=0}^{n} (n-k) \, x^{k} = \sum_{k=0}^{n-2} (n-k) \, x^{k} + x^{n-1}$$
then
\begin{align}
\sum_{k=0}^{n-2} (n-k) \, x^{k} &= \sum_{k=0}^{n} k \, x^{n-k} - x^{n-1} 
= \frac{(2x-1) x^{n-1} - (n+1) x + n}{(x-1)^{2}}
\end{align}
for the case of $x=2$ this becomes
$$ \sum_{k=0}^{n-2} (n-k) \, 2^{k} = 3 \cdot 2^{n-1} - n - 2.$$
