Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Hint: start from the binomial theorem and differentiate accordingly... – J. M. Jul 24 '11 at 16:20
possible duplicate of How to compute the following formula – Aryabhata Jul 24 '11 at 18:45
@Aryabhata:Thanks for that pointing out to that question. – Quixotic Jul 24 '11 at 20:31
up vote 8 down vote accepted

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,

share|cite|improve this answer
+1 & accepted;Excellent answer! precisely the one hyper-linked. – Quixotic Jul 24 '11 at 20:30

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 $2$.

share|cite|improve this answer
+1 for "derivative with respect to 2" (and clever answer). :) – Ilmari Karonen Jul 24 '11 at 17:20
Yes, lotta fun differentiating with respect to constants, but best to do it before they completely settle down. :) – paul garrett Jul 24 '11 at 20:01
Failing to differentiate between variables and constants invariably and constantly leads to trouble. – joriki Jul 24 '11 at 21:08

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.