Summation of $\sum_{i=1}^n i2^i.$ An intermediate step in a problem I was working on was to find a closed form for the sum $$\sum_{i=1}^n i2^i.$$
WolframAlpha returns $2^{n+1}(n-1) + 2$, but didn't provide any step-by-step solution.  I would usually provide progress that I made, but I wasn't sure how to proceed(although I solved the problem by bashing out small values).

To find the formula on your own.  Write
$$
S = \sum_{i=1}^n i2^i.
\tag{1}$$
Then
$$
2S = \sum_{i=1}^n i2^{i+1} = \sum_{i=2}^{n+1} (i-1)2^i
\tag{2}$$
Subtract (1) from (2)
$$
S = 2S - S = \sum_{i=2}^{n+1} (i-1)2^i - \sum_{i=1}^n i2^i 
\\
= n2^{n+1} +\sum_{i=2}^n (-1)2^i -1\cdot 2^1
$$
The sum remaining is  a geometric series, I assume you already know how to do it
$$
S =   n2^{n+1} - \left(\frac{2^{n+1}-2^2}{2-1}\right) - 2= 
(n-1)2^{n+1} + 2
$$
