Question

$\frac{1}{(3^2-1)}+\frac{1}{(5^2-1)}+\frac{1}{(7^2-1)}+\dots \dots .+\frac{1}{({201}^2-1)}$ is equal to

• A. $\frac{101}{404}$
• B. $\frac{99}{400}$
• C. $\frac{25}{101}$
• D. $\frac{101}{408}$

Answer 1

The correct option is C

$\frac{25}{101}$

Explanation for the correct option:

Finding the value of the given expression:

Given sum $\frac{1}{(3^2-1)}+\frac{1}{(5^2-1)}+\frac{1}{(7^2-1)}+\dots \dots .+\frac{1}{({201}^2-1)}$

Since, each term in given expression can be written in the form of $a_n=\frac{1}{{\left(2n+1\right)}^2-1}$ so, Taking Sum of this expression we get

$$\begin{gathered} \sum _{n=1}^{100}=\frac{1}{{\left(2n+1\right)}^2-1}=\frac{1}{4n^2+4n+1-1}[{(a+b)}^2=a^2+2ab+b^2] \\ =\sum _{n=1}^{100}\frac{1}{2n(2n+2)} \\ =\frac{1}{2}\sum _{n=1}^{100}\frac{2}{2n(2n+2)} \\ =\frac{1}{2}\sum _{n=1}^{100}\frac{(2n+2)-2n}{2n(2n+2)} \\ =\frac{1}{2}\sum _{n=1}^{100}\left(\frac{1}{2n}-\frac{1}{(2n+2)}\right) \\ =\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{8}\right)+.....+\left(\frac{1}{200}-\frac{1}{202}\right) \\ =\frac{1}{2}\left(\frac{1}{2}-\frac{1}{202}\right) \\ =\frac{100}{2\times 202} \\ =\frac{25}{101} \end{gathered}$$

Hence the correct answer is option (C).