If {i^2} = -1,then the value of sum_{n=1}^{20 | vedclass

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(C) The given expression is a geometric series: $\sum_{n=1}^{200} {i^n} = i + i^2 + i^3 + \dots + i^{200}$.Here,the first term $a = i$,the common rati

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(C) The given expression is a geometric series: $\sum_{n=1}^{200} {i^n} = i + i^2 + i^3 + \dots + i^{200}$.Here,the first term $a = i$,the common ratio $r = i$,and the number of terms $n = 200$.The sum of a geometric series is given by $S_n = \frac{a(1 - r^n)}{1 - r}$.Substituting the values: $S_{200} = \frac{i(1 - i^{200})}{1 - i}$.Since $i^4 = 1$,we have $i^{200} = (i^4)^{50} = 1^{50} = 1$.Therefore,$S_{200} = \frac{i(1 - 1)}{1 - i} = \frac{0}{1 - i} = 0$.

If ${i^2} = -1$,then the value of $\sum_{n=1}^{200} {i^n}$ is

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