Q. The value of sum $\sum_{n = 1}^{13}{\left( i^{n} + i^{n + 1} \right),\ }\ where\ i = \sqrt{- 1},\$equals
A) $i$
B) $i - 1\$
C) $- i$
D) 0
Explanation
(b)
As sum of any four consecutive powers of iota is zero
$\therefore\ \ \sum_{n = 1}^{13}{\left( i^{n} + i^{n + 1} \right) = \left( i + i^{2} + ... + i^{13} \right) + (i^{2} + i^{3} + ... + i^{14})}$
$= i + i^{2} = i - 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.