Finding closed form for $\sum_{k=1}^n k2^{k-1}$ I am trying to use the perturbation method to find a closed form for:
$$
S_n = \sum_{k=1}^n k2^{k-1}
$$
This is what I’ve tried so far:
$$
S_n + (n+1)2^n = 1 + \sum_{k=2}^{n+1} k2^{k-1}
$$
$$
S_n + (n+1)2^n = 1 + \sum_{k=1}^{n} (k+1)2^{k}
$$
$$
S_n + (n+1)2^n = 1 + \sum_{k=0}^{n-1} k2^{k-1}
$$
$$
S_n + (n+1)2^n = 1 + 0 + n2^{n-1} + \sum_{k=1}^{n} k2^{k-1}
$$
But I don’t think it is right since it would result in a $S_n - S_n = ...$ 
Could you please help me?
Thank you very much.

Note the mistake: $\sum_{k=1}^{n} (k+1)2^{k}\ne \sum_{k=0}^{n-1} k2^{k-1}$.
Here is the right way:
$$\begin{align}S_n + (n+1)2^n &= 1 + \sum_{k=1}^{n} (k+1)2^{k}=\\
&=1+\sum_{k=0}^{n-1} (k+2)2^{k+1}=\\
&=1+\color{blue}{\sum_{k=0}^{n-1} k\cdot 2^{k+1}}+\color{red}{\sum_{k=0}^{n-1}2^{k+2}}=\\
&=1+\color{blue}{\sum_{k=0}^n k\cdot 2^{k+1}-n\cdot 2^{n+1}}+\color{red}{4(2^n-1)}=\\
&=1+4\sum_{k=0}^nk\cdot 2^{k-1}-n\cdot 2^{n+1}+4(2^n-1)=\\
&=1+4S_n-n\cdot 2^{n+1}+4(2^n-1) \Rightarrow \\
3S_n&=(n+1)2^n-1+n\cdot 2^{n+1}-4(2^n-1)=\\
&=3n\cdot 2^n-3\cdot 2^n+3 \Rightarrow \\
S_n&=n\cdot 2^n-2^n+1.\end{align}$$
