Question

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

• A. 99/199
• B. 1/199
• C. 100/201
• D. 1/201

Answer 1

The correct option is C 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 $\left(\frac{1}{35}\right)=\frac{1}{(7\times 5)}=\frac{1}{2}[\frac{1}{5}-\frac{1}{7}]$
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100th term of the series $\left(\frac{1}{199\times 201}\right)=\frac{1}{2}[\frac{1}{199}-\frac{1}{201}]$
$⟹N=\frac{1}{2}[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots +\frac{1}{199}-\frac{1}{201}]$
$⟹N=\frac{1}{2}[1-\frac{1}{201}]=\frac{200}{(2\times 201)}=\frac{100}{201}$