4-1.Complex numbers
easy

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

A

$50$

B

$-50$

C

$0$

D

$100$

Solution

(c) $\sum\limits_{n = 1}^{200} {{i^n} = i + {i^2} + {i^3} + …. + {i^{200}} = \frac{{i(1 – {i^{200}})}}{{1 – i}}} $ (since G.P.)
$ = \frac{{i(1 – 1)}}{{1 – i}} = 0$.

Standard 11
Mathematics

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