If {i^2} = -1,then the sum i + {i^2} + {i^3} | vedclass

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(D) The given series is a geometric progression with first term $a = i$ and common ratio $r = i$.The sum of the first $n$ terms of a geometric progres

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(D) The given series is a geometric progression with first term $a = i$ and common ratio $r = i$.The sum of the first $n$ terms of a geometric progression is given by $S_n = \frac{a(1 - r^n)}{1 - r}$.For $n = 1000$ terms,$S_{1000} = \frac{i(1 - i^{1000})}{1 - i}$.Since $i^4 = 1$,we have $i^{1000} = (i^4)^{250} = 1^{250} = 1$.Substituting this into the sum formula: $S_{1000} = \frac{i(1 - 1)}{1 - i} = \frac{i(0)}{1 - i} = 0$.

If ${i^2} = -1$,then the sum $i + {i^2} + {i^3} + \dots$ up to $1000$ terms is equal to

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