$ \left( \frac{1}{3} + \frac{4}{7} \right) + \left( \frac{1}{3^2} + \frac{1}{3} \times \frac{4}{7} + \frac{4^2}{7^2} \right) + \left( \frac{1}{3^3} + \frac{1}{3^2} \times \frac{4}{7} + \frac{1}{3} \times \frac{4^2}{7^2} + \frac{4^3}{7^3} \right) + \ldots \text{ upto infinite terms is equal to }$

Let $ a = \frac{4}{7},; b = \frac{1}{3} $

Multiply $ N $ and $ D $ by $ (a - b) = \frac{4}{7} - \frac{1}{3} = \frac{5}{21} $

$ \frac{1}{a - b} \left[ (a^2 - b^2) + (a^3 - b^3) + (a^4 - b^4) + \ldots \right] $

$ = \frac{1}{a - b} \left[ \frac{a^2}{1 - a} - \frac{b^2}{1 - b} \right] $

$ = \frac{21}{5} \left[ \frac{16/49}{1 - 4/7} - \frac{1/9}{1 - 1/3} \right] $

$ = \frac{21}{5} \left[ \frac{16/49}{3/7} - \frac{1/9}{2/3} \right] $

$ = \frac{21}{5} \left[ \frac{16}{21} - \frac{1}{6} \right] $

$ = \frac{21}{5} \cdot \frac{96 - 21}{126} $

$ = \frac{21}{5} \cdot \frac{75}{126} $

$ = \frac{15}{6} = \frac{5}{2} $