Q. Let $S = \frac{8}{5} + \frac{16}{15} + ... + \frac{128}{2^{18} + 1},$ then
A) $S = \frac{1088}{545}$
B) $S = \frac{545}{1088}$
C) $S = \frac{1056}{545}$
D) $S = \frac{545}{1056}$
Explanation
(a)
$S = \sum_{r = 1}^{16n}\left( \frac{8r}{4r^{4} + 1} \right)$
$= 2\ \sum_{r = 1}^{16n}\left( \frac{1}{2r^{2} - 2r + 1} - \frac{1}{2r^{2} + 2r + 1} \right)$
$= 2\left( 1 - \frac{1}{5} + \frac{1}{5} - \frac{1}{13} + \frac{1}{13} + ... + \frac{1}{481} - \frac{1}{545} \right)$
$= 2\left( 1 - \frac{1}{545} \right) = \frac{1088}{545}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.