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!
 A: Hint
Consider $$\sum_{i=0}^p i x^i=x\sum_{i=0}^p i x^{i-1}=x\left(\sum_{i=0}^p  x^i \right)'$$
A: 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
$$ \sum_{k=0}^{n-2} (n-k) \, 2^{k} = 3 \cdot 2^{n-1} - n - 2.$$
A: $$\sum_{j=1}^{n-2}j\;2^j=2+2\times 2^2+3\times 2^3+\cdots+(n-2)2^{n-2}$$$$=[2+2^2+2^3+\cdots+2^{n-2}]+[2^2+2^3+\cdots+2^{n-2}]+[2^3+\cdots+2^{n-2}]+\cdots+2^{n-2}$$$$=2(2^{n-2}-1)+2^2(2^{n-3}-1)+2^3(2^{n-4}-1)+\cdots+2^{n-2}(2^{n-(n-1)}-1)$$ $$=(2^{n-1}-2)+(2^{n-1}-2^2)+(2^{n-1}-2^3)+\cdots+(2^{n-1}-2^{n-2})$$ $$=(n-2)2^{n-1}-(2+2^2+\cdots+2^{n-2})=(n-2)2^{n-1}-2(2^{n-2}-1)$$ $$=(n-2)2^{n-1}-2^{n-1}+2=(n-3)2^{n-1}+2$$
