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Question:

The value of 1 + $\left(1+\frac{1}{3}\right)\frac{1}{4}$ + $\left(1+\frac{1}{3}+\frac{1}{9}\right)\frac{1}{16}$ + $\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)\frac{1}{64}$ + ..., is

Explanation:

Given, 1 + $\left(1+\frac{1}{3}\right)\frac{1}{4}$ + $\left(1+\frac{1}{3}+\frac{1}{9}\right)\frac{1}{16}$ + $\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)\frac{1}{64}$ + ..., 

= 1 + $\left(\frac{1}{4}+\frac{1}{3}\times \frac{1}{4}\right)$ + $\left(\frac{1}{16}+\frac{1}{3}\times \frac{1}{16}+\frac{1}{9}\times \frac{1}{16}\right)$ + $\left(\frac{1}{64}+\frac{1}{3}\times \frac{1}{64}+\frac{1}{9}\times \frac{1}{64}+\frac{1}{27}\times \frac{1}{64}\right)$ + ...

= 1 + $\left(\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+...\right)$ + $\frac{1}{3}\left(\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+...\right)$ + $\frac{1}{9}\left(\frac{1}{16}+\frac{1}{64}+\frac{1}{256}+...\right)$ + $\frac{1}{27}\left(\frac{1}{64}+\frac{1}{256}+\frac{1}{1024}...\right)$ + ...

= 1 + $\left(\frac{1/4}{1-1/4}\right)$ + $\frac{1}{3}\left(\frac{1/4}{1-1/4}\right)$ + $\frac{1}{9}\left(\frac{1/16}{1-1/4}\right)$ + $\frac{1}{27}\left(\frac{1/64}{1-1/4}\right)$ + ...

= 1 + $\frac{1}{3}$ + $\frac{1}{9}$ + $\frac{1}{9}\times \frac{1}{12}$ + $\frac{1}{9}\times \frac{1}{144}$ + ...

= 1 + $\frac{1}{3}$ + $\frac{1}{9}\left(1+\frac{1}{12}+\frac{1}{144}+...\right)$

= 1 + $\frac{1}{3}$ + $\frac{1}{9}\times \frac{1}{1-1/12}$

= 1 + $\frac{1}{3}$ + $\frac{4}{33}$

= $\frac{33+11+4}{33}$

= $\frac{48}{33}$

= $\frac{16}{11}$

Hence, option (c).