# 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 $$a=i$$.

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

$$\sum\limits_{i=1}^{n}{i\cdot 2^i}= \sum\limits_{i=1}^{n}{\sum_{k=1}^i 2^i}=\sum_{k=1}^n\sum_{i=k}^n2^i$$ $$=\sum_{k=1}^n2^k\sum_{i=k}^n2^{i-k}\stackrel{i=l+k}{=}\sum_{k=1}^n2^k\sum_{l=0}^{n-k}2^{l}$$ $$= \sum_{k=1}^n2^k(2^{n-k+1}-1) = n2^{n+1}-\sum_{k=1}^n2^k$$ $$=n2^{n+1}-2(2^n-1)= 2(n-1)2^n+2$$

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 $$2f(2).$$

• Thank you! I never would have thought of doing that.
– Jim
Commented Mar 29, 2020 at 14:55

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?

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:

$$g_{n+1}(2)=\frac{2^{n+2}(2(n+1)-(n+1)-1)+2}{1^2}=n\,2^{n+2}+2$$