Question

The sum of the series $\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\dots \dots \dots$ terms, is

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $\frac{3}{4}$
• D. $\frac{5}{6}$

Answer 1

Answer: (A)

$S=\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+\dots \dots \infty$
$=\frac{1}{2}[\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{9}\right)+\dots \dots \infty =\frac{1}{2}$.