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).
 A: 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
$$
S =   n2^{n+1} - \left(\frac{2^{n+1}-2^2}{2-1}\right) - 2= 
(n-1)2^{n+1} + 2
$$
A: Let the sum equals $S$.$$\big(1\cdot x+2\cdot x^2+\cdots +n\cdot x^n\big)+\big(n\cdot x^n+(n-1)x^{n-1}+\cdots +x\big)=(n+1)(1+x+\cdots+x^n-1)$$
Hence, $2S=(n+1)(1+x+\cdots +x^n)-(n+1)=(n+1)\cdot \frac{x^{n+1}-1}{x-1}-(n+1) $ and finally we get $$S=\frac{n+1}{2}(\frac{x^{n+1}-1}{x-1})-\frac{n+1}{2}$$
If you put $x=2$ you will get the required sum.
A: I assume you know the formula for the geometric sum:
$$\sum_{k=0}^n x^k=\frac{x^{n+1}-1}{x-1}$$
It can be shown by multiplication by $x-1$ and cancellation of terms.
Your sum is a little more complicated.
$$\sum_{k=1}^n kx^k$$
However, we can write it as a triangular array. Let's use $n=5$ for clarity.
$$\begin{array}( x^5 & + & x^4 & + & x^3 & + & x^2 & + & x  \\  x^5 & + & x^4 & + & x^3 & + & x^2  \\  x^5 & + & x^4 & + & x^3  \\  x^5 & + & x^4 \\ x^5 \end{array}$$
Can you see the general pattern? Maybe this will help.
A: Hint:
$$2S_4-S_4=(1\cdot2^2+2\cdot2^3+3\cdot2^4+4\cdot2^5)-(1\cdot2^1+2\cdot2^2+3\cdot2^3+4\cdot2^4)
\\=4\cdot2^5-(2^1+2^2+2^3+2^4)=3\cdot2^5+2.$$
You can easily generalize.
