Question

If $S=1+i^2+i^4+\dots +i^{2n},$ then which of the following is/are true

• A. $S=0,$ when $n$ is odd
• B. $S=0,$ when $n$ is even
• C. $S=1,$ when $n$ is even
• D. $S=1,$ when $n$ is odd

Answer 1

Answer: (A), (C)

The correct options are
A $S=0,$ when $n$ is odd
C $S=1,$ when $n$ is even

$S=1+i^2+i^4+\dots +i^{2n}$

This is clearly a $G.P.$ having $n+1$ terms with first term $a=1$ and common ratio $r=i^2.$

$\therefore S=\frac{a(1-r^{n+1})}{1-r}\Rightarrow S=\frac{1(1-(i^2{)}^{n+1})}{1-i^2}\Rightarrow S=\frac{1-(-1{)}^{n+1}}{2}$
$\Rightarrow S=\{\begin{array}{c}\frac{1-1}{2}=0,\text{when}n\text{is odd}\\ \frac{1+1}{2}=1,\text{when}n\text{is even}\end{array}$

Alternate solution:
We know $i^{4m}=1$ and $i^{4m+2}=-1$
$\therefore S=(1-1)+(1-1)+...+i^{2n}$
If $n$ is even then $2n$ will be of the form$4m$
So, $S=(1-1)+(1-1)+(\mathrm{1...}-1)+1$
$\therefore S=1$
If $n$ is odd then $2n$ will be of the form$4m+2$
So, $S=(1-1)+(1-1)+\mathrm{1...}+(1-1)$
$\therefore S=0$