The value of $1+i^2+i^4+i^6+....+i^{2n}$ is

The correct option is B Cannot be determined

$i^1=i$

$i^2=-1$

$i^3=-i$

$i^4=1$

Other powers of i can be easily found out.

$i^5=i^2\times i^3=-1\times -i=i$

Similarly

$i^6=i^4\times i^2=1\times -1=-1$

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

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

In the question

if n is odd the last term will be -1 and hence the summation is:

$1+i^2+i^4+i^6+....+i^{2n}=1+(-1)+1(-1)+\mathrm{1......}(-1)=0$

if n is even the last term will be 1 and hence the summation is:

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

So the answer is $0$ or $1$ depending on whether n is odd or even.

Unless $n$ is specified exact answer cannot be determined.