# Solve the following

Question:

If $P^{2}=-1$, then the sum $i+\beta^{2}+\beta^{3}+\ldots$ upto 1000 terms is equal to

(a) 1

(b) −1

(c) i

(d) 0

Solution:

(d) 0

$i+i^{2}+i^{3}+i^{4} \ldots i^{1000}$

$i+i^{2}+i^{3}+i^{4} \quad\left[\because i^{2}=-1, i^{3}=-i\right.$ and $\left.i^{4}=1\right]$

$=i-1-i+1$

$=0$

Similarly, the sum of the next four terms of the series will be equal to 0 . This is because the powers of $i$ follow a cyclicity of 4 .

Hence, the sum of all terms, till 1000 , will be zero.

$i+i^{2}+i^{3}+i^{4} \ldots i^{1000}=0$