Question

Sum of infinite terms of series $\frac{1}{7}+\frac{2}{7^2}+\frac{1}{7^3}+\frac{2}{7^4}+....$ is

• A. $\frac{1}{6}$
• B. $\frac{3}{16}$
• C. $\frac{1}{16}$
• D. $\frac{7}{16}$

Answer 1

The correct option is B $\frac{3}{16}$

Now,

$\frac{1}{7}+\frac{2}{7^2}+\frac{1}{7^3}+\frac{2}{7^4}+....$

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

$=\frac{\frac{1}{7}}{1-\frac{1}{7^2}}+2\frac{\frac{1}{7^2}}{1-\frac{1}{7^2}}$

$=\frac{7}{48}+\frac{1}{24}$

$=\frac{9}{48}$

$=\frac{3}{16}$.