# 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.

• Easier to divide by $5^{4}$ and you have $\sum_{i=0}^{N-4} 5^i$. Jan 31, 2016 at 16:54
• I'm not really sure why that would work, but my issue is still that I don't get how to solve geometric series, so I still am a bit lost.
– Jacc
Jan 31, 2016 at 17:00
• For $\sum_{k=m}^n r^k$, the general formula is(for $r \neq 1$) $$(\text{first summand})\times\frac{1-r^{\text{number of summands}}}{1-r} = r^m \frac{1-r^{n-m+1}}{1-r}$$ (and, if $r = 1$, it is simply $\text{number of summands} = (n-m+1)$). But you original thought is right -- it is not the fastest way, but it'll work. Jan 31, 2016 at 17:07

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: $$S = 5^4 \times {5^{n-3}-1\over4}$$

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 $$-1$$: $$\sum_{i=m}^{N} r^i = \frac{r^m - r^{N+1}}{1-r}$$