Q:

Sum rule and $\sum_{k=1}^n (n-k)2^{k-1}$

I am trying to obtain a closed form for $\sum_{k=1}^n (n-k)2^{k-1}$ by a combinatorial argument. Unfortunately, nothing seems to strike. If I write this as $n\sum 2^{k-1}-\sum k2^{k-1}$ then I do have the first term $n\sum 2^{k-1}=n2^{n+1}-n2^n-n$ but am unable to make sense of the second term.
Can someone help please?

A:

Hint:
How many binary sequences of length $n$ contain at least two 0's, with the second 0 occurring in position $n-k+1$?

