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Question: The value of \(\sum_{r = 1}^{102}i^{r}\)is –...

The value of ∑r=1102ir\sum_{r = 1}^{102}i^{r}is –

A

0

B

I

C

– I

D

–1

Answer

0

Explanation

Solution

Sol. ∑r=1102ir\sum_{r = 1}^{102}i^{r}= i + i2 + i3 ……..+i102

= i(i102−1)i−1\frac{i(i^{102} - 1)}{i - 1}= i(1−1)i−1\frac{i(1 - 1)}{i - 1}= 0