Question

Find the value of $1+i^2+i^4+i^6+...+i^{2n}$ where $i=\sqrt{-1},n\in N$

Answer 1

$\because 1+i^2+i^4+i^6+...+i^{2n}=1-1+1-1+...{(-1)}^n$

Case I:If $n$ is odd

$1+i^2+i^4+i^6+...+i^{2n}=1-1+1-1+...+1-1=0$

Case II:If $n$ is even

$1+i^2+i^4+i^6+...+i^{2n}=1-1+1-1+...+1=1$