JEE PYQ: Sequence And Series Question 6
Question 6 - 2021 (18 Mar Shift 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
(1) $\frac{101}{404}$
(2) $\frac{25}{101}$
(3) $\frac{101}{408}$
(4) $\frac{99}{400}$
Show Answer
Answer: (2)
Solution
$T_n = \frac{1}{(2n+1)^2 - 1} = \frac{1}{4n(n+1)} = \frac{1}{4}\left(\frac{1}{n} - \frac{1}{n+1}\right)$
$S = \frac{1}{4}\left(1 - \frac{1}{101}\right) = \frac{1}{4} \cdot \frac{100}{101} = \frac{25}{101}$
BETA
