Q35 / 90 · JEE Main 2022 (25 Jul Shift 2) · Mathematics

The sum $\sum_{n = 1}^{21} \frac{3}{(4 n - 1) (4 n + 3)}$ is equal to

A
$\frac{7}{87}$
Correct answer
B
$\frac{7}{29}$
C
$\frac{14}{87}$
D
$\frac{21}{29}$

Solution

$\sum_{n = 1}^{21} \frac{3}{(4 n - 1) (4 n + 3)} = \frac{3}{4} \sum_{n = 1}^{21} (\frac{1}{4 n - 1} - \frac{1}{4 n + 3})$
$= \frac{3}{4} [(\frac{1}{3} - \frac{1}{7}) + (\frac{1}{7} - \frac{1}{11}) + \dots . + (\frac{1}{83} - \frac{1}{87})]$
$= \frac{3}{4} [\frac{1}{3} - \frac{1}{87}] = \frac{3}{4} \frac{84}{261} = \frac{7}{29}$

Hence, the correct answer is option: 'B'.

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