# Math Help - Geometric Series

1. ## Geometric Series

Find the number of terms and the sum of the geometric series below.

$x+x^3+x^5+...+x^25$

$n=13$
$S_{13}=\frac{x(1-x^{26})}{1-x^2}$

$S_{13}=\frac{x(1-x^{26})}{1-x^2}$
but not
$S_{13}=\frac{x(x^{26}-1)}{x^2-1}$
?

2. Originally Posted by SengNee
Find the number of terms and the sum of the geometric series below.

$x+x^3+x^5+...+x^25$

$n=13$
$S_{13}=\frac{x(1-x^{26})}{1-x^2}$

$S_{13}=\frac{x(1-x^{26})}{1-x^2}$
but not
$S_{13}=\frac{x(x^{26}-1)}{x^2-1}$
?
As far as I can remember(and I did series a long time ago), it does not matter how they are arranged. You should arrive at the same answer I think(EDIT: Yes you will).

If r < 1 then we use one of the arrangements, and if r > 1 then we use the other. (EDIT: Only to make things easier)

3. Originally Posted by SengNee
$S_{13}=\frac{x(1-x^{26})}{1-x^2}$
$S_{13}=\frac{x(x^{26}-1)}{x^2-1}$