Solution to the geometric progression starting from an arbitrary index? We know that:
$$
\sum_{n=0}^{n-1}r^n = \frac{r^n -1}{r-1}
$$
What about when the starting index is an arbitrary $n_0$? Is the following correct?:
$$
\sum_{n=n_0}^{n-1}r^n = \sum_{n=0}^{n-1}r^{n+n_0} = \sum_{n=0}^{n-1}r^{n}r^{n_0} = r^{n_0}\sum_{n=0}^{n-1}r^{n} = r^{n_0}\left(\frac{r^n -1}{r-1}\right)
$$
I can't find any explicit solution to this.
 A: Using $$
\sum_{n=0}^{n-1}r^n = \frac{r^n -1}{r-1}$$
Try computing  
$$
\begin{align} \sum_{n_0}^{n-1}r^n \quad 
& = \sum_{n= 0}^{n-1}r^{n} - \sum_{n=0}^{n_0-1}r^{n} \\ \\
& = \frac{r^n -1}{r-1} - \frac{r^{n_0} - 1}{r-1} \\ \\ 
& = \frac{r^n - r^{n_0}}{r-1} \\ \\
& = r^{n_0}\left(\frac{r^{n - n_0}-1}{r-1}\right)
\end{align}
$$
A: I think this step is wrong $$\sum_{n=n_0}^{n-1}r^n = \sum_{n=0}^{n-1}r^{n+n_0}$$
A: Let $n-n_o=m$. Then 
\begin{align}
\sum_{k=n_o}^{n} r^k
=
&
\big(
r^{n_o}+r^{n_o+1}+r^{n_o+2}+\ldots +r^{n-1}+r^{n}
\big)
\\
=
&
\big(
r^{n_o}+r^{n_o+1}+r^{n_o+2}+\ldots +r^{n_o+m-1}+r^{n_o+m}
\big)
\\
=
&
\big(
r^{n_o}\cdot 1+r^{n_o}\cdot r^1+r^{n_o}\cdot r^2+\ldots +r^{n_o}\cdot r^{+m-1}+r^{n_o}\cdot r^m
\big)
\\
=
&
r^{n_o}\big( 1+ r^1+r^2+\ldots +r^{n-n_o-1}+r^{n-n_o}\big)
\\
=
&
r^{n_o}\cdot \sum_{k=0}^{n-n_o}r^k
\end{align}
A: Here is an easy way to remember the sum formula. The sum of a geometric series is $${{\rm first} - {\rm blast}\over 1 - {\rm common\ ratio}},$$
where ``blast'' is the term that is one beyond the last.  This works no matter where the sum begins or ends.  If the common ratio has absolute value less than one and you are summing to $\infty$, then the blast term is $0$.
