I am trying to prove
$$ \sum_{k=1}^n k2^{k-1} = 1+(n-1)2^n $$
I proved the base case with $n = 1$. I am having trouble proving the induction step.
I know I need to prove for $n = n +1$ so I got
$$ \sum_{k=1}^n k2^{k-1}+(k+1)2^{(k+1)-1} = 1+[(n+1)-1]2^{n+1} $$
suppose $n = i$
$$ 1 + [(i+1)-1] 2^{i+1} $$
I am not sure if I am on the right path and how to simplify from here onwards? Could someone point me in the right direction?
$$ \sum_{k=1}^{n+1}k2^{k-1}= \biggl(\sum_{k=1}^{n}k2^{k-1}\biggr)+(n+1)2^{n+1-1}= 1+(n-1)2^{n}+(n+1)2^n $$ Now we can finish it up: $$ 1+(n-1)2^{n}+(n+1)2^n=1+(n-1+n+1)2^n= 1+2n\cdot 2^n=1+n2^{n+1} $$ and the statement you had to prove is exactly $$ \sum_{k=1}^{n+1}k2^{k-1}=1+n2^{n+1} $$ (just substitute $n$ with $n+1$ in the original formula).
You had $k$ in the term you pushed out of the summation, which is wrong.
