Question:
Find the sum ( $\mathrm{i}+\mathrm{i}^{2}+\mathrm{i}^{3}+\mathrm{i}^{4}+\ldots .$ up to 400 terms)., where $\mathrm{n} \mathrm{N}$.
Solution:
We have, $\mathrm{i}+\mathrm{i}^{2}+\mathrm{i}^{3}+\mathrm{i}^{4}+\ldots$ up to 400 terms
We know that given series is GP where a=i , r = i and n = 400
Thus, $S=\frac{a\left(1-r^{n}\right)}{1-r}$
$=\frac{\mathrm{i}\left(1-(\mathrm{i})^{400}\right)}{1-\mathrm{i}}$
$=\frac{\mathrm{i}\left(1-\left(\mathrm{i}^{4}\right)^{100}\right)}{1-\mathrm{i}}$
$=\frac{\mathrm{i}\left(1-1^{100}\right)}{1-\mathrm{i}}\left[\because \mathrm{i}^{4}=1\right]$
$=\frac{\mathrm{i}(1-1)}{1-\mathrm{i}}=0$