Mathematical Induction (summation): $\sum^n_{k=1} k2^k =(n-1)(2^{n+1})+2$ I am stuck on this question from the IB Cambridge HL math text book about Mathematical induction. I am sorry about the bad formatting I am new and have no idea how to write the summation sign.
Using mathematical induction prove that the 
$$\sum^n_{k=1} k2^k =(n-1)(2^{n+1})+2$$ 
[correction made]
I tried solving it and got stuck on the let $n=k+1$ part 
So first I made $n=1$ and both sides equaled to $2$ 
then assume $n=k$ and got an expression which I don't know how to write here because of the formatting
then $n=K+1$ 
Thanks again 

$\sum_{k=1}^{n}k2^{k}=2+(4+4)+(8+8+8)+...+(2^{n}+2^{n}+...+2^{n})$
$=(2+4+8+...+2^{n-1}+2^{n})+(4+8+...+2^{n-1}+2^{n})+(8+...+2^{n-1}+2^{n})+...+(2^{n-2}+2^{n-1}+2^{n})+(2^{n-1}+2^{n})+2^{n}$
$2(1+2+4+...+2^{n-2}+2^{n-1})+4(1+2+4+...+2^{n-3}+2^{n-2})+8(1+...+2^{n-3})+...+2^{n-2}(1+2+4)+2^{n-1}(1+2)+2^{n}$
$=2(2^{n}-1)+4(2^{n-1}-1)+8(2^{n-2}-1)+...+2^{n-2}(2^{3}-1)+2^{n-1}(2^{2}-1)+2^{n}$
$=(n-1)2^{n+1}+2^{n}-2-4-...-2^{n-1}=(n-1)2^{n+1}+2^{n}-2(2^{n-1}-1)$
$=(n-1)2^{n+1}+2$
This should be the formula. Now we prove this by induction. Both sides are $2$ at $n=1$. Assume it is true for $n\ge1$ and we show it for $n+1$.
$\sum_{k=1}^{n+1}k2^{k}=\sum_{k=1}^{n}k2^{k}+(n+1)2^{n+1}=(n-1)2^{n+1}+2+(n+1)2^{n+1}$
$=(2n)2^{n+1}+2=n2^{n+2}+2=((n+1)-1)2^{(n+1)+1}+2$.
