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(D) We know that the sum of four consecutive powers of $i$ is zero,i.e.,$i^n+i^{n+1}+i^{n+2}+i^{n+3}=0$ for any integer $n$. The given series is $S =

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$-i$

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(D) We know that the sum of four consecutive powers of $i$ is zero,i.e.,$i^n+i^{n+1}+i^{n+2}+i^{n+3}=0$ for any integer $n$. The given series is $S = i^2+i^3+i^4+\ldots+i^{4000}$. This series contains $4000-2+1 = 3999$ terms. We can group these terms into sets of four. Since $3999 = 4 	imes 999 + 3$,there are $999$ groups of four consecutive powers of $i$ (each summing to $0$) and $3$ remaining terms. The sum of the first $3996$ terms is $0$. Thus,$S = i^{3997} + i^{3998} + i^{3999} + i^{4000} = 0$ is not the correct approach here; rather,we observe the sequence $i^2, i^3, i^4, \ldots, i^{4000}$. The sum is $\sum_{k=2}^{4000} i^k = \frac{i^2(1-i^{3999})}{1-i} = \frac{-1(1-i^3)}{1-i} = \frac{-1(1-(-i))}{1-i} = \frac{-(1+i)}{1-i} = \frac{-(1+i)(1+i)}{(1-i)(1+i)} = \frac{-(1+2i+i^2)}{1-i^2} = \frac{-(1+2i-1)}{1+1} = \frac{-2i}{2} = -i$.

$i^2+i^3+\ldots+i^{4000}=$

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