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Value of sum(k=1)^(100)(i^(k!)+omega^(k!...

Value of `sum_(k=1)^(100)(i^(k!)+omega^(k!))`, where `i=sqrt(-1)` and `omega` is complex cube root of unity, is :

A

`190 + omega`

B

`192 + omega^(2)`

C

`190 + i`

D

192 + i

Text Solution

Verified by Experts

The correct Answer is:
D
`sum_(k =1)^(100) i^(k!) + sum_(k = 1)^(100)omega^(k!)`
`sum_(k=1)^(100)i^(k!) = i^(1!) + i^(2!) + i^(3!)+ i^(4!) + ....i^(100!)`
`= I - 1 + i^(6) + 1 + 1 + 1 +…. + 1`
= I - 2 + 97 = I + 95.
`sum_(k=1)^(100) omega^(k!) = omega^(1!) + omega^(2!) + omega^(3!) + omega^(4!) + omega^(100!)`
`= omega + omega^(2) + 1 + 1 +1 +...+ 1 = 97`
sum = i + 95 + 97 = i + 192
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