Question

If $i^2=-1$ then the value of $\sum _{n=1}^{200}{i}^{2n}$ is:

• A. $50$
• B. $10$
• C. $0$
• D. $100$

Answer 1

Answer: (C)

$0$

As $i^2=-1$

$\Rightarrow i^{2n}=(-1{)}^n$

For $n=$ odd, $i^{2n}=-1$

For $n=$ even, $i^{2n}=1$

There are equal number of even and odds in between $1$ to $200$

$\Rightarrow {\sum }_{n=1}^{200}=-1+1-1+1-.......=0$