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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

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


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marked as duplicate by apnorton, Kirill, Rafflesia arnoldii, Hanul Jeon, John ZHANG Nov 8 '14 at 2:14

This question was marked as an exact duplicate of an existing 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

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.

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