Calculate $\sum_{k=1}^n k2^{n-k}$ I want to Calculate $\sum_{k=1}^n k2^{n-k}$. Here's my attempt:
$$\begin{align} 
&\sum_{k=1}^n k2^{n-k} = \sum_{k=1}^n k2^{n+1}2^{-k-1} =2^{n+1}\sum_{k=1}^n k2^{-k-1} & &\text{(1)} \\
&k2^{-k-1} = -\frac{d}{dk}2^{-k} & &\text{(2)}
\end{align}$$ 
Plugging $(2)$ to $(1)$, 
$$\begin{aligned} 
2^{n+1}\sum_{k=1}^n k2^{-k-1}  &= 2^{n+1}\sum_{k=1}^n-\frac{d}{dk}2^{-k} \\ 
&= -2^{n+1}\sum_{k=1}^n\frac{d}{dk}2^{-k} \\ 
&= -2^{n+1}\frac{d}{dk}\sum_{k=1}^n2^{-k} \\ 
&= -2^{n+1}\frac{d}{dk}(1-2^{-n}) \\
&= -2^{n+1}\cdot 0 \\
&= 0 \end{aligned}$$ 
Obviously, I don't know much about summation and derivative. Can you please let me know where I am doing wrong and give me a hint for this question? Thanks! 

Let $S=\sum_{k=1}^n k 2^{n-k}$
$ S = 2^{n-1} + 2(2^{n-2}) + 3(2^{n-3}) + \cdots + n(2^0)$
$ {S \over 2} = \quad\quad\quad1(2^{n-2}) + 2(2^{n-3}) + \cdots + (n-1)(2^0) + n(2^{-1})$ 
$\begin{align} (1-{1\over2})S 
&= (2^{n-1} + 2^{n-2} + 2^{n-3} + \cdots + 1) - n(2^{-1})\cr
{S\over2}&= (2^n-1) - n(2^{-1})\cr\cr
S &= 2(2^n-1) - n
\end{align}$
