Non-infinite geometric sum; does not start at 0 or 1 It's bee a long time since I've worked with sums and series, so even simple examples like this one are giving me trouble:
$\sum_{i=4}^N \left(5\right)^i$
Can I get some guidance on series like this? I'm finding different methods online but not sure which to use. I know that starting at a non-zero number also changes things.
My original thought was to do (sum from 0 to N of 5^i) - (sum from 0 to 3 of 5^i) but I'm not sure that's right.
 A: The problem asks for a closed-form solution to:
$$\sum_{i=4}^{N} 5^i = 5^4 + 5^5 + ... + 5^N$$
The OP's original intuition was correct:
$$\sum_{i=4}^{N} = \sum_{i=0}^{N} 5^i - \sum_{i=0}^{3} 5^i$$
More generally, for summing a geometric series starting at an arbitrary index $m$:
$$
\sum_{i=m}^{N} r^i = \sum_{i=0}^{N} r^i - \sum_{i=0}^{m-1} r^i \\
$$
To get a closed form for the above expression, let's start with the closed-form equation for a geometric series:
$$
\sum_{i=0}^{N} r^i = \frac{r^{N+1}-1}{r-1}
$$
So:
$$
\begin{align*}
\sum_{i=m}^{N} r^i &= \sum_{i=0}^{N} r^i - \sum_{i=0}^{m-1} r^i \\
  &= (\frac{r^{N+1}-1}{r-1}) - (\frac{r^{m-1+1}-1}{r-1}) \\
&= \frac{r^{N+1} - r^m}{r-1}
\end{align*}
$$
An alternative and equivalent form can be found if we multiply the top and bottom by $-1$:
$$
\sum_{i=m}^{N} r^i = \frac{r^m - r^{N+1}}{1-r}
$$
A: Let $S = a + ar + ar^2 + ar^3 ...$
Then $S-Sr = (a + ar + ar^2 + ar^3 ... ar^n) - (ar + ar^2 + ar^3 + ar^4 ... ar^{n+1}) = a - ar^{n+1}$
Factoring out an S we have $S(1-r) = a-ar^{n+1}$
Finally, $$S = {(a - ar^{n+1})\over(1-r)}$$
In your case, you are trying to find $5^4 + 5^5 + 5^6 ... 5^n$
You can factor out a $5^4$ to get $5^4(1 + 5 + 5^2 ... + 5^{n-4})$
Plugging in corresponding values of $a$ and $r$ into the equation above we have:
$$S = 5^4 \times {5^{n-3}-1\over4} $$
