Question

If $i^{2}=-1$ and $\displaystyle \sum _{r = 1}^{n}\left(i\right)^{r},\forall n\in N,$ is a non-zero real number, then $n$ can be

• A. $100$
• B. $201$
• C. $302$
• D. $403$

Answer 1

Answer: (D)

$i^{p}+i^{p+1}+i^{p+2}+i^{p+3}=0, \forall p \in N$
So $\sum_{r=1}^{n}(i)^{r}=\left\{\begin{array}{cc}0 & \text { for } n=4 k \\ i & \text { for } n=4 k+1 \\ i-1 & \text { for } n=4 k+2 \\ -1 & \text { for } n=4 k+3\end{array} \forall k \in N\right.$
Hence, $n$ can be $403$ for the sum to be non-zero real.