How do you get from $\sum\limits_{i=0}^{n}{i\cdot 2^i}$ to $2(n-1)\cdot 2^n+2$? I'm currently trying to expand the following sum and have simplified it down using the steps below:
$$
\sum_{i=1}^{n+1}{i\cdot2^i}\\
\sum_{i=0}^{n}{(i+1)\cdot2^{(i+1)}}\\
\sum_{i=0}^{n}{i\cdot2^{i+1}}+\sum_{i=0}^{n}{2^{i+1}}\\
\sum_{i=0}^{n}{2\cdot i\cdot2^{i}}+\sum_{i=0}^{n}{2\cdot 2^{i}}\\
2\sum_{i=0}^{n}{i\cdot2^{i}}+2\sum_{i=0}^{n}{2^{i}}\\
$$
I know I can use the formula for a geometric series on the right sum to get $2\sum_{i=0}^{n}{2^{i}}=2(2\cdot 2^n-1)$
But I'm unsure of how to simplify the left sum since $a=i$.
 A: You can calculate the sum directly.
Noting that your sum actually starts at $i=1$ because the summand for $i=0$ is equal to $0$, you get
$$\sum\limits_{i=1}^{n}{i\cdot 2^i}= \sum\limits_{i=1}^{n}{\sum_{k=1}^i 2^i}=\sum_{k=1}^n\sum_{i=k}^n2^i$$
$$=\sum_{k=1}^n2^k\sum_{i=k}^n2^{i-k}\stackrel{i=l+k}{=}\sum_{k=1}^n2^k\sum_{l=0}^{n-k}2^{l}$$
$$= \sum_{k=1}^n2^k(2^{n-k+1}-1) = n2^{n+1}-\sum_{k=1}^n2^k$$
$$=n2^{n+1}-2(2^n-1)= 2(n-1)2^n+2$$
A: You have proved that $\sum_{i=1}^{n+1}{i\cdot2^i}=2\sum_{i=0}^{n}{i\cdot2^i}+2\sum_{i=1}^{n}{\cdot2^i}$. Denoting the last term by $A$ we get $\sum_{i=1}^{n}{i\cdot2^i}+(n+1)2^{n+1}=2\sum_{i=1}^{n}{i\cdot2^i}+A$. This can be written as $\sum_{i=1}^{n}{i\cdot2^i}=(n+1)2^{n+1}-A$ (by transferring the left side to the right side and $A$ to the left side). Can you finish?
A: On a more general point of view when seeing things like $i\,2^i$ you have to think derivative.
Indeed if we call $\quad\displaystyle f_n(x)=\sum\limits_{i=0}^n x^i=\dfrac{x^{n+1}-1}{x-1}\quad$ and $\quad\displaystyle g_n(x)=\sum\limits_{i=0}^n i\,x^i$
Then ${f_n}'(x)=\sum\limits_{i=1}^n i\,x^{i-1}=\frac 1x\sum\limits_{i=1}^n i\,x^i=\frac 1xg_n(x)$ 
$$g_n(x)=x\,{f_n}'(x)=\dfrac{x^{n+1}(nx-n-1)+x}{(x-1)^2}$$
You are interested in: 
$g_{n+1}(2)=\frac{2^{n+2}(2(n+1)-(n+1)-1)+2}{1^2}=n\,2^{n+2}+2$
A: Note that
$$
\sum_{k=0}^nk2^k=2\sum_{k=0}^nk2^{k-1}
$$
then define
$$
f(x):=\sum_{k=0}^nkx^{k-1}
$$
and
$$
g(x):=\sum_{k=0}^nx^{k}=\frac{x^{n+1}-x}{1-x}\;.
$$
Clearly
$$
f(x)=g'(x)=\frac{\left((n+1)x^n-1\right)(1-x)+x^{n+1}-x}{(1-x)^2}.
$$
What we where searching for is then 
$
2f(2).
$
