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Q.
If $i^2=-1$, then the value of $\displaystyle\sum_{n=1}^{200} i^n$ is
Complex Numbers and Quadratic Equations
Solution:
$\displaystyle\sum_{n=1}^{200} i^n=i+i^2+i^3+\ldots+i^{200}$
$ =\frac{i\left(1-i^{200}\right)}{1-i}$ (using sum formula of G.P.)
$ =\frac{i(1-1)}{1-i}=0$