Q. ${(2}^{3n} - 1)$ will be divisible by $(\forall n \in N)$
A) 25
B) 8
C) 7
D) 3
Explanation
(c)
$2^{3n} - 1 = (2^{3})^{n} - 1$
$= 8^{n} - 1 = (1 + 7)^{n} - 1$
$= 1 + \ ^{n}C_{1}7 + \ ^{n}C_{2}7^{2} +$…$+ \ ^{n}C_{n}7^{n} - 1$
$= 7\lbrack\ ^{n}C_{1} + \ ^{n}C_{2}7 +$…$+ \ ^{n}C_{n}7^{n - 1}\rbrack$
$\therefore 2^{3n} - 1$ is divisible by 7
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.