Question

$1+i^2+i^4\dots .+i^{2n}is$

• A. Positive
• B. Negative
• C. Zero
• D. Cannot be determined

Answer 1

The correct option is D

Cannot be determined

Explanation for the correct option:

Determine the value of given expression taking even and odd values of $n$

As we know,

$i^{2n}=-1$ when $n$ is odd that is $n=1,3,5,...$

and $i^{2n}=1$when $n$ is even that is $n=2,4,6...$

Let given sum $S=1+i^2+i^4\dots .+i^{2n}is$

Put $n=oddnumber$

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

Put $n=evennumber$

$\Rightarrow S=1+i^2+i^4+i^6+i^8+....+i^{2n}=1-1+1-1+1=1$

Therefore, The value will vary according to the value of $n$.

So, the value of $1+i^2+i^4\dots .+i^{2n}$ can not be determined unless the value of $n$ is specified as even or odd.

Hence, Option ‘D’ is Correct.