TITLE: A "fast" way for computing $\sum\limits_{n=1}^{100} n\times 2^n $
QUESTION [5 upvotes]: How to compute 'z', where $\displaystyle z = \sum_{n=1}^{100} n\times 2^n$ ?

The answer is of the form $99 \times 2^{101} + 2$, I need a fast approach as this problem is supposed to be solved under a minute.

REPLY [11 votes]: If you define $y=\sum_{n=1}^{100}x^n=\frac{x^{101}-1}{x-1}$, then $z=x\frac{dy}{dx}$ evaluated at $x=2$.  I love taking the derivative with respect to $2$.

REPLY [8 votes]: We can actually solve this problem without calculus.  (See this answer.)
Full Solution:  Lets find a general formula for the following sum for any $r,m$: $$S_{m}=\sum_{n=1}^{m}nr^{n}.$$ 
This can be derived in a similar manner to the formula for the geometric series. Notice that $$S_{m}-rS_{m}=-mr^{m+1}+\sum_{n=1}^{m}r^{n}$$
$$=-mr^{m+1}+\frac{r-r^{m+1}}{1-r}=\frac{mr^{m+2}-(m+1)r^{m+1}+r}{1-r}.$$  Hence $$S_m = \frac{mr^{m+2}-(m+1)r^{m+1}+r}{(1-r)^2}.$$
This equality holds for any $r$, so by letting $r=2$ we are able to conclude
$$\sum_{n=1}^m n2^n = m2^{m+2}-(m+1)2^{m+1}+2=2^{m+1}(m-1)+2.$$
Hope that helps,