Question

The sum of the infinite series:
$\frac{1}{10}+\frac{2}{{10}^2}+\frac{3}{{10}^3}+.........+\frac{n}{{10}^n}+.....$

• A. $\frac{1}{9}$
• B. $\frac{10}{81}$
• C. $\frac{1}{8}$
• D. $\frac{17}{22}$

Answer 1

The correct option is B $\frac{10}{81}$

For sum of any AGP, $a+(a+d)r+(a+2d)r^2+(a+3d)r^3+.........$

Sum= $\frac{a}{1-r}+\frac{rd}{(1-r{)}^2}$

In our problem, we have $a=\frac{1}{10},r=\frac{1}{10},d=\frac{1}{10}$

Sum=$\frac{\frac{1}{10}}{1-\frac{1}{10}}+\frac{\left(\frac{1}{10}\right)\left(\frac{1}{10}\right)}{{(1-\frac{1}{10})}^2}$

$=\frac{1}{9}+\frac{1}{81}=\frac{10}{81}$