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Question
Write the sum of the series \[i + i^2 + i^3 + . . . .\] upto 1000 terms.
Solution
We know that, \[i + i^2 + i^3 + i^4 = i - 1 - i + 1 = 0\]
\[\therefore i + i^2 + i^3 + . . . . + i^{1000} \]
\[ = \left( i + i^2 + i^3 + i^4 \right) + \left( i^5 + i^6 + i^7 + i^8 \right) + . . . + \left( i^{997} + i^{998} + i^{999} + i^{1000} \right)\]
\[ = \left( i + i^2 + i^3 + i^4 \right) + \left( i^4 i + i^4 i^2 + i^4 i^3 + i^4 i^4 \right) + . . . + \left[ \left( i^4 \right)^{249} i + \left( i^4 \right)^{249} i^2 + \left( i^4 \right)^{249} i^3 + \left( i^4 \right)^{249} i^4 \right]\]
\[ = \left( i + i^2 + i^3 + i^4 \right) + \left( i + i^2 + i^3 + i^4 \right) + . . . + \left( i + i^2 + i^3 + i^4 \right)\]
\[ = 0\]
Thus, the sum of the series
\[i + i^2 + i^3 + . . . .\] upto 1000 terms is 0.
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