If i=sqrt{-1},then sum_{n=2}^{30} i^n+sum_{n= | vedclass

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(B) We know that the sum of four consecutive powers of $i$ is $i^k + i^{k+1} + i^{k+2} + i^{k+3} = i^k(1 + i - 1 - i) = 0$. For the first sum: $\sum_{

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$-1$

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(B) We know that the sum of four consecutive powers of $i$ is $i^k + i^{k+1} + i^{k+2} + i^{k+3} = i^k(1 + i - 1 - i) = 0$. For the first sum: $\sum_{n=2}^{30} i^n = i^2 + i^3 + \dots + i^{30}$. The number of terms is $30 - 2 + 1 = 29$. Since $29 = 4 	imes 7 + 1$,the sum is $i^2 + (i^3 + i^4 + i^5 + i^6) + \dots + (i^{27} + i^{28} + i^{29} + i^{30}) = -1 + 0 = -1$. For the second sum: $\sum_{n=30}^{65} i^{n+3} = i^{33} + i^{34} + \dots + i^{68}$. The number of terms is $68 - 33 + 1 = 36$. Since $36$ is a multiple of $4$,the sum of these $36$ consecutive powers of $i$ is $0$. Thus,the total sum is $-1 + 0 = -1$.

If $i=\sqrt{-1}$,then $\sum_{n=2}^{30} i^n+\sum_{n=30}^{65} i^{n+3}=$

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