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