How to sum a series with different index

Question

I am new to this site so not sure if this type of question is appropriate. I know that the sum of a geometric series can be written like this:

$$ S = \sum ^n_{k=1} a^k = \frac{a^1 - a^n}{1-a} $$

How does this change however if the power of each term is not the same, for instance $2k+1$ or $2k-1$ etc?

$$ S = \sum^n_{k=1} a^{2k-1} \stackrel{?}{=} \frac{a^1 - a^{2n-1}}{1-a} $$

Kind regards!

score 1 · Answer 1 · answered Dec 14 '17 at 13:50

1

Hint: $a^{2k-1}=a^{-1}\cdot(a^2)^k$

answered Dec 14 '17 at 13:50

MrYouMath

score 1 · Accepted Answer · answered Dec 14 '17 at 13:51

1

in this Case we get $$\sum_{k=1}^na^{2k-1}=\frac{a^{2n+2}-a^2}{a(a^2-1)}=\frac{a^{2n+1}-a}{a^2-1}$$

answered Dec 14 '17 at 13:51

Thank you for your answer, in fact working it through, I get exactly -1 * your answer, $(a-a^{n+1})/1-a^2$? – user513534 Dec 14 '17 at 14:15
multiply numerator and denominator by $-1$ – Dr. Sonnhard Graubner Dec 14 '17 at 14:17

2 Answers2