If i=sqrt{-1},then 1+i^2+i^4+i^6+ldots+i^{202 | vedclass

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(C) The given series is $S = 1+i^2+i^4+i^6+\ldots+i^{2024}$.We know that $i^2 = -1$,$i^4 = 1$,$i^6 = -1$,and so on.This is a geometric progression wit

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$1$

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(C) The given series is $S = 1+i^2+i^4+i^6+\ldots+i^{2024}$.We know that $i^2 = -1$,$i^4 = 1$,$i^6 = -1$,and so on.This is a geometric progression with first term $a=1$,common ratio $r=i^2=-1$,and the number of terms $n = \frac{2024-0}{2} + 1 = 1013$.The sum of a geometric series is $S_n = \frac{a(r^n - 1)}{r - 1}$.Substituting the values: $S = \frac{1((-1)^{1013} - 1)}{-1 - 1} = \frac{-1 - 1}{-2} = \frac{-2}{-2} = 1$.

If $i=\sqrt{-1}$,then $1+i^2+i^4+i^6+\ldots+i^{2024} = $

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