Write the sum of the series

We know that,

$i+i^{2}+i^{3}+i^{4}=i-1-i+1=0$

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

$=\left(i+i^{2}+i^{3}+i^{4}\right)+\left(i^{5}+i^{6}+i^{7}+i^{8}\right)+\ldots+\left(i^{997}+i^{998}+i^{999}+i^{1000}\right)$

$=\left(i+i^{2}+i^{3}+i^{4}\right)+\left(i^{4} i+i^{4} i^{2}+i^{4} i^{3}+i^{4} i^{4}\right)+\ldots+\left[\left(i^{4}\right)^{249} i+\left(i^{4}\right)^{249} i^{2}+\left(i^{4}\right)^{249} i^{3}+\left(i^{4}\right)^{249} i^{4}\right]$

$=\left(i+i^{2}+i^{3}+i^{4}\right)+\left(i+i^{2}+i^{3}+i^{4}\right)+\ldots+\left(i+i^{2}+i^{3}+i^{4}\right)$

$=0$

Thus, the sum of the series $i+i^{2}+i^{3}+\ldots$ upto 1000 terms is 0 .