Infinite sum for a geometric series. I am asked to find the sum to the following infinite geometric series:
$\sum_{n=1}^{\infty}\frac{(2)(3^{n+1})}{5^n}$
I then factor out the 2 and one 3 from the $3^{n+1}$ and get:
$\sum_{n=1}^{\infty}6 (\frac{3}{5})^n$
this results in a = 6 and r = $\frac{3}{5}$
and therefore I found the sum to be:
$\frac {6}{1 - \frac{3}{5}} = 15$
I thought this to be correct, however in the answer key for some reason it redefines the limit of the sum such that:
$\sum_{n=1}^{\infty}6 (\frac{3}{5})^n = -6 + \sum_{n=0}^{\infty}6 (\frac{3}{5})^n $
which returns the sum of 9 instead. Why does he does he redefine the bounds of the summation like this? Are these answers essentially the same?

The geometric series formula you have is for a series starting at $n = 0$. If you want to apply that formula, you have to add and subtract that initial term, which is $6$.
That is, you have, whether you like it or not,
$$\sum_{n=1}^{\infty}6 (\frac{3}{5})^n$$
Your formula only works if the sum starts at $n = 0$. So, we fix that:
$$\sum_{n=1}^{\infty}6 (\frac{3}{5})^n = -6 + 6 + \sum_{n=1}^{\infty}6 (\frac{3}{5})^n = -6 + \sum_{n=0}^{\infty}6 (\frac{3}{5})^n $$
