The finite form of the summation was written as such, but they state the general form of a finitie series is to the n-1 power. So I would rewrite this as, if I followed the form that was mentioned

$\displaystyle \sum_{n=4}^{20}(\frac{1}{3})^{n-1}$

Now If I follow this, the first term I get is $\displaystyle (\frac{1}{3})^3$

Following the form for a finite series partial sum is

$\displaystyle \frac{\frac{1}{3}^3(1-\frac{1}{3}^{20})}{1-\frac{1}{3}}$

However, this is not quite right as provided by the book or other solutions I've seen my friends do of this problem, when I do the way I found is more accurate for me

$\displaystyle \sum_{n=4}^{20}(\frac{1}{3})^n$

Where I look at the finite series partial sum form as this, the first term unbroken to help simplifying exponents

$\displaystyle \frac{ar^n-ar^{n+1}}{1-r}$

Where $\displaystyle ar^n$ is the first term in the series and $\displaystyle ar^{n+1}$ is the last term in the series + 1, works the same for the general case of the nth term

$\displaystyle \frac{(\frac{1}{3}^4-\frac{1}{3}^{21})}{1-\frac{1}{3}}$

Similarily

$\displaystyle \frac{\frac{1}{3}^4(1-\frac{1}{3}^{20})}{1-\frac{1}{3}}$

Which is the correct term before further simplifiction, so whats the deal with the form they mention and other sources I have seen explaining the geometric series? Confuses me at times. Sorry for the lengthy post, I know how MHF want poster to include all information