Question

Find the sum of the first $27$ terms of the geometric series $\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+.....$

Answer 1

To find the sum of first $27$ terms of the geometric series $\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+........$, we have to first find the common ratio $r$.

In the given geometric series, the first term is $a_1=\frac{1}{9}$ and the second term is $a_2=\frac{1}{27}$ and so on.

We find the common ratio $r$ by dividing the second term by first term as shown below:

$r=\frac{\frac{1}{27}}{\frac{1}{9}}=\frac{1}{27}\times \frac{9}{1}=\frac{1}{3}<1$

We know that the sum of an geometric series with first term $a$ and common ratio $r$ is $S_n=\frac{a(1-r^n)}{1-r}$ if $r<1$

Now, to find the sum of first $27$ terms, substitute $a=\frac{1}{9},r=\frac{1}{3}$ and $n=27$ in $S_n=\frac{a(1-r^n)}{1-r}$ as follows:

$S_{27}=\frac{\frac{1}{9}[1-{\left(\frac{1}{3}\right)}^{27}]}{1-\frac{1}{3}}=\frac{\frac{1}{9}[1-{\left(\frac{1}{3}\right)}^{27}]}{\frac{3-1}{3}}=\frac{\frac{1}{9}[1-{\left(\frac{1}{3}\right)}^{27}]}{\frac{2}{3}}=\frac{1}{9}\times \frac{3}{2}[1-{\left(\frac{1}{3}\right)}^{27}]=\frac{1}{6}[1-{\left(\frac{1}{3}\right)}^{27}]$

Hence the sum is $\frac{1}{6}[1-{\left(\frac{1}{3}\right)}^{27}]$.