Question

Calculate the sum of series $\frac{1}{7}+\frac{2}{7^2}+\frac{3}{7^3}+.........$ upto infinity.

• A. $\frac{1}{36}$
• B. $\frac{14}{36}$
• C. $\frac{7}{36}$
• D. $\frac{7}{64}$

Answer 1

The correct option is C $\frac{7}{36}$

Series:

$S=\frac{1}{7}+\frac{2}{7^2}+\frac{3}{7^3}+...\infty$ ........(1)

$\frac{1}{7}.S=\frac{1}{7^2}+\frac{2}{7^3}+\frac{3}{7^4}+...\infty$ ........(2)

$\therefore$ eq (1) - eq(2)

$\Rightarrow S(1-\frac{1}{7})=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...\infty$

$\Rightarrow S\left(\frac{6}{7}\right)=\frac{\frac{1}{7}}{1-\frac{1}{7}}$

$\Rightarrow S\left(\frac{6}{7}\right)=\frac{1}{6}$

$\therefore S=\frac{7}{36}$