So Here's a few problems where I have to find the nth partial sum:

Given the equation $\displaystyle S_n = a\frac{1-r^{(n+1)}}{1-r} $

The sum of a Geometric series for a sequence $\displaystyle A_n$ such that it follows the general pattern $\displaystyle A_n = ar^k $ starting at n=1. a is a constant, and

|r| < 1

If we let n go towards infinity, then $\displaystyle S_n = \frac{a}{1-r} $

However, before I can begin dealing with infinity, I can't get the general equation for the sum up to the nth term. What am I doing wrong?

The Sequence is: $\displaystyle 2, \frac{2}{5^2}, \frac{2}{5^3}, .... , \frac{2}{5^{(k-1)}} $

$\displaystyle S(\frac{2}{5^{(n-1)}}) = 2(\frac{1}{5})^{(n-1)} = 10(\frac{1}{5})^n $

if we let a = 10 and r = 1/5, then from Sn I got

$\displaystyle 10\frac{(1-(\frac{1}{5})^{k+1})}{1-\frac{1}{5}}=10\frac{(1-(\frac{1}{5})^k\frac{1}{5}}{\frac{4}{5}}=$ $\displaystyle \frac{25}{2}(\frac{5}{5}-(\frac{1}{5})^k\frac{1}{5})=\frac{5}{2}(5-(\frac{1}{5})^k)$

That's what I got. However the way it is done is:

$\displaystyle \frac{2}{5^{(k-1)}} = 2(\frac{1}{5})^{(k-1)}$

let a = 2 and let r= 1/5

$\displaystyle S(2(\frac{1}{5})^{(k-1)}=2\frac{\(1-(\frac{1}{5})^{k-1+1}}{1-\frac{1}{5}}= 2\frac{\(1-(\frac{1}{5})^{k})}{1-\frac{1}{5}}= 2\frac{\(1-(\frac{1}{5})^{k}}{\frac{4}{5}}=\frac{5}{2}(1-(\frac{1}{5})^{k}$

Manual calculation by hand shows the second method to be the correct one. While I accept it, I can't see what's wrong with the first.