JEE Main 2021 (Online) 18th March Morning ShiftQuestion 1
$\frac{1}{3^{2}-1}+\frac{1}{5^{2}-1}+\frac{1}{7^{2}-1}+\ldots+\frac{1}{(201)^{2}-1}$ is equal to :
Sequence and Series · Telescoping ($V_n$ Method)
Solution
Formulas used. (1) $\frac1{k^2-1}=\frac12\left(\frac1{k-1}-\frac1{k+1}\right)$Step 1 - Split. For $k=3,5,7,\dots,201$: $$\frac1{k^2-1}=\frac12\left(\frac1{k-1}-\frac1{k+1}\right).$$ Step 2 - Telescope. $$\frac12\left[\left(\frac12-\frac14\right)+\left(\frac14-\frac16\right)+\cdots+\left(\frac1{200}-\frac1{202}\right)\right]=\frac12\left(\frac12-\frac1{202}\right).$$ Step 3. $$\frac12\cdot\frac{101-1}{202}=\frac{50}{202}=\frac{25}{101}.$$ Answer: (A)