The value of the series 1 + i^2 + i^4 + i^6 + | vedclass

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(D) The given series is $S = 1 + i^2 + i^4 + i^6 + ..... + i^{2n}$.Since $i^2 = -1$,$i^4 = 1$,$i^6 = -1$,and so on,the series becomes $S = 1 - 1 + 1 -

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Cannot be determined

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(D) The given series is $S = 1 + i^2 + i^4 + i^6 + ..... + i^{2n}$.Since $i^2 = -1$,$i^4 = 1$,$i^6 = -1$,and so on,the series becomes $S = 1 - 1 + 1 - 1 + ..... + (-1)^n$.This is a finite geometric series with $n+1$ terms,where the first term $a = 1$ and the common ratio $r = i^2 = -1$.The sum of a geometric series is $S = \frac{a(1 - r^{n+1})}{1 - r}$.Substituting the values,$S = \frac{1(1 - (-1)^{n+1})}{1 - (-1)} = \frac{1 - (-1)^{n+1}}{2}$.If $n$ is even,$n+1$ is odd,so $S = \frac{1 - (-1)}{2} = \frac{2}{2} = 1$.If $n$ is odd,$n+1$ is even,so $S = \frac{1 - 1}{2} = 0$.Since the value depends on the parity of $n$,it cannot be determined without knowing $n$.

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

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