Question

The value of $\frac{1}{i}+\frac{1}{i^{2}}+\frac{1}{i^{3}}+\cdots+\frac{1}{i^{102}}$ is

• A. $-1 - i$
• B. $1 + i$
• C. $1 - i$
• D. $1 + 2i$

Answer 1

Answer: (A)

$-1 - i$

Answer 2

$\frac{1}{i}+\frac{1}{i^{2}}+\frac{1}{i^{3}}+\ldots+\frac{1}{i^{102}}$
\therefore\, S_{n}=\frac{\frac{1}{i}\left\\{1-\left(\frac{1}{i}\right)^{102}\right\\}}{1-\left(\frac{1}{i}\right)}=\frac{(1+1)}{i-1}
$\begin{pmatrix}\because i^{4}=1 \\\ \text { and } i^{2}=-1\end{pmatrix}$
$=\frac{2(i+1)}{i^{2}-1}=-(i+1)=-1-i$