The formula for the nth partial sum of a geometric sequence is,

\(\displaystyle{S}_{{n}}={\frac{{{a}_{{1}}{\left({1}-{r}^{{n}}\right)}}}{{{1}-{r}}}},\ {r}\ne{1}\)

where, \(\displaystyle{S}_{{n}}\) is the sum of GP with n terms

\(\displaystyle{a}_{{1}}\) is the first term

r is the common ratio

n is the number of terms

The sum of the geometric series given is,

\(\displaystyle{\frac{{{1}}}{{{3}}}}+{\frac{{{2}}}{{{9}}}}+{\frac{{{4}}}{{{27}}}}+{\frac{{{8}}}{{{21}}}}+{\frac{{{16}}}{{{243}}}}\)

There are 5 terms in the series and the common ratio of the series is found by dividing any term by the previous term. Suppose, divide the second term by the first term and the common ratio is obtained as,

\(\displaystyle{r}={\frac{{{2}}}{{{9}}}}\div{\frac{{{1}}}{{{3}}}}\)

\(\displaystyle={\frac{{{2}}}{{{3}}}}\)

Therefore, r is not equal to 1 and the sum of the given series is,

\(S_5=\frac{\frac{1}{3}(1-(\frac{2}{3})^5)}{(1-\frac{2}{3})}\)