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_(r=4)^(100)i^(r!)+prod_(r=1)^(101)i^(r)=a+ib, " where " i=sqrt(-1)`, then a+75b, is

A

11

B

22

C

33

D

44

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The correct Answer is:

B

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