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Sum of four consecutive powers of i(iota...

Sum of four consecutive powers of i(iota) is zero.
i.e.,`i^(n)+i^(n+1)+i^(n+2)+i^(n+3)=0,forall n in I.`
If `sum_(n=1)^(25)i^(n!)=a+ib, " where " i=sqrt(-1)`, then a-b, is

A

prime number

B

even number

C

composite number

D

perfect number

Text Solution

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The correct Answer is:
A
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