# How do I find the sum of the geometric series 8 + 4 + 2 + 1?

Now, this is called a finite sum, because there are a countable set of terms to be added. The first term, ${a}_{1} = 8$ and the common ratio is $\frac{1}{2}$ or .5. The sum is calculated by finding: S_n= frac{a_1(1-R^n)}{(1-r) = $\frac{8 \left(1 - {\left(\frac{1}{2}\right)}^{4}\right)}{\setminus} \left(1 - \frac{1}{2}\right)$ = $\frac{8 \left(1 - \frac{1}{16}\right)}{1 - \left(\frac{1}{2}\right)}$ =$8 \frac{\left(\frac{15}{16}\right)}{\frac{1}{2}}$ = $\left(\frac{8}{1}\right) \left(\frac{15}{16}\right) \left(\frac{2}{1}\right)$ = 15.
$\frac{{a}_{1} \left({r}^{n} - 1\right)}{r - 1}$. Try it on a different problem!