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Question: The value of \(\sum_{n = 0}^{100}i^{n!}\) equals (where i = \(\sqrt{- 1}\))...

The value of ∑n=0100in!\sum_{n = 0}^{100}i^{n!} equals (where i = −1\sqrt{- 1})

A

–1

B

I

C

2i + 95

D

97 + i

Answer

2i + 95

Explanation

Solution

Sol. S = ∑n=0100(i)n\sum_{n = 0}^{100}{(i)^{\begin{matrix} n \end{matrix}}}

S = (i)0+(i)1+(i)2+...(i)^{\begin{matrix} 0 \end{matrix}} + (i)^{\begin{matrix} 1 \end{matrix}} + (i)^{\begin{matrix} 2 \end{matrix}} + ...

= i + i – 1 + i6 + i24 + (i)5+(i)6(i)^{\begin{matrix} 5 \end{matrix}} + (i)^{\begin{matrix} 6 \end{matrix}} + .....+ (i)100(i)^{\begin{matrix} 100 \end{matrix}}

= 95 + 2i