Question

Find the sum of infinity of the series $1+\frac{2.1}{3}+\frac{3.1}{3^2}+\frac{4.1}{3^3}+$............. .

Answer 1

$S=1+\frac{2.1}{3}+\frac{3.1}{3^2}+\frac{4.1}{3^3}....\left(1\right)S.\frac{1}{3}=\frac{1}{3}+\frac{2.1}{3^2}+\frac{3.1}{3^3}+\frac{4.1}{3^4}+....\left(2\right)$

Equation $\left(1\right)-\left(2\right)$

$S-S.\frac{1}{3}=1+(\frac{2.1}{3}-\frac{1}{3})+(\frac{3.1}{2^2}-\frac{2.1}{3^2})+(\frac{4.1}{3^3}-\frac{3.1}{3^3})+....\frac{2}{3}S=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....\frac{2}{3}S=\frac{1}{1-\frac{1}{3}}$(sum of infinite term of G.P)

$\frac{2}{3}S=\frac{3}{2}\therefore S=\frac{9}{4}$

The sum of the series is $\frac{9}{4}$