Question

Find the sum of the series $N=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}.........100$ terms

• A. $\frac{99}{199}$
• B. $\frac{1}{199}$
• C. $\frac{100}{201}$
• D. $\frac{1}{201}$
• E. $\frac{1}{199}$

Answer 1

The correct option is C $\frac{100}{201}$

First term of the series $\left(\frac{1}{3}\right)=\frac{1}{(3\times 1)}=\frac{1}{2}[1-\frac{1}{3}]$
Second term of the series $\left(\frac{1}{15}\right)=\frac{1}{(3\times 5)}=\frac{1}{2}[\frac{1}{3}-\frac{1}{5}]$
Third term of the series $\left(\frac{1}{35}\right)=\frac{1}{(7\times 5)}=\frac{1}{2}[\frac{1}{5}-\frac{1}{7}]$
..............................................................................
100th term of the series $(\frac{1}{199}\times 201)=\frac{1}{2}[\frac{1}{199}-\frac{1}{201}]$
$\frac{1}{2}[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}...................+\frac{1}{199}-\frac{1}{201}]$
$\frac{1}{2}[1-\frac{1}{201}]=\frac{200}{(2\times 201)}=\frac{100}{201}$