Question 20.

The given sequence can be written as:

$1\left(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...\right) + \frac{1}{3}\left(\frac{1}{4} + \frac{1}{16} + ...\right) + \frac{1}{9}\left(\frac{1}{16} + \frac{1}{64} + ...\right) + ...$

We know that the sum of an infinite G.P. is $\frac{a}{1-r}$, where a is the first term and r is the common ratio.

=> The first term = $\frac{1}{1-\frac{1}{4}} = \frac{4}{3}$

=> The second term = $\frac{1}{3}\left(\frac{\left(\frac{1}{4}\right)}{1-\left(\frac{1}{4}\right)}\right) = \frac{1}{9}$

=> The third term = $\frac{1}{9}\left(\frac{\left(\frac{1}{16}\right)}{1-\left(\frac{1}{4}\right)}\right) = \frac{1}{108}$

Observing these three terms, we see that they are in G.P. with a common ratio of $\frac{1}{12}$

=> Sum of this infinite G.P. = $\frac{\left(\frac{4}{3}\right)}{1-\left(\frac{1}{12}\right)} = \frac{16}{11}$