Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This question already has an answer here:

I would like to evaluate the sum of the following geometric progression

$1 -2 + 2^{2} - 2^{3} + ...+(-1)^{n}2^{n}$

Would the following proposed solution be on the right lines?

$a = 1$ (Being the first term)

$r = -2$ (Being the common ratio)

$n = n + 1$ (The number of terms we want to consider in this case)

The formula to evaluate the sum of a geometric progression being:

$$\frac{1 - r^n}{1 - r}.$$

Therefore, plugging in the values above

$$\frac{1 - (-2)^{n+1}}{3}.$$


share|improve this question

marked as duplicate by anorton, Kirill, Fundamental, tetori, John ZHANG Nov 8 '14 at 2:14

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Your proposed solution is unreadable. What do you mean by $a$ and $r$? It looks like you're referring to some particular result but are leaving it to the reader to guess which it is and how your letters connect to it. –  Henning Makholm Oct 27 '12 at 15:08
The parentheses are wrong.. it should be $\frac{1 - (-2)^{n+1}}{3}$ –  Cocopuffs Oct 27 '12 at 15:11
Cheers..Thanks for the heads up on brackets..Still trying to get used to the editing on site –  bosra Oct 27 '12 at 15:14
Writing "$n = n + 1$" is always dubious, regardless of the further content. In this case you are referring to two different variables it seems, so just write something like $m = n + 1$ then. –  TMM Oct 27 '12 at 17:53

1 Answer 1

It would help to recall the derivation of the formula for GP, where you took sum of GP as S, multiplied it with "r" and wrote in shifted places so that all except first and last term of S and r*S would match and then finally subtracted then to get (1-r)*S on LHS. Hence, as long as r is not 1, the formula would hold, also in your case.

share|improve this answer

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