Question

Sum of infinite terms of the series $\frac{1.2}{3}+\frac{2.3}{3^2}+\frac{3.4}{3^3}+\frac{4.5}{3^4}+....\infty$ is

• A. $\frac{9}{4}$
• B. $\frac{1}{9}$
• C. $\frac{5}{4}$
• D. $\frac{1}{4}$

Answer 1

Answer: (A)

$\frac{9}{4}$

$S=\frac{1.2}{3}+\frac{2.3}{3^2}+\frac{3.4}{3^3}+\frac{4.5}{3^4}+....$

$\Rightarrow \frac{1}{3}S=0+\frac{1.2}{3^2}+\frac{2.3}{3^3}+\frac{3.4}{3^4}+....$
On Subtraction both sides, $\frac{2}{3}S=\frac{2}{3}+\frac{4}{3^2}+\frac{6}{3^3}+\frac{8}{3^4}...$
Multiplying $\frac{1}{3}$ on both the sides, we get

$\frac{2}{9}S=\frac{2}{3^2}+\frac{4}{3^3}+\frac{6}{3^4}+\frac{8}{3^5}....$
On Subtraction $\frac{4}{9}S=\frac{2}{3}+\frac{2}{3^2}+\frac{2}{3^3}+\frac{2}{3^4}...=\frac{\frac{2}{3}}{1-\frac{1}{3}}=1$

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