The value of `i^(1+3+5++(2n+1))` is, If n is odd.

Options

A. `i`
B. `1`
C. `-1`
D. `-i`

Answer

B

Explanation

Let `z = i^([1+3+5+......+(2n+1)])`
Now , `1+3+5+…..+(2n+1)` are `n+1` terms of an A.P whose sum is
`(n+1)/(2)[1+2n +1]=(n+1)^(2)`
Hence, `z=i^((n+1))`. Now put n = 1,2,3,4,5....
`n =1,z=i^(4) = 1,n=2, z = i^(9) = i`
`n=3,z=i^(16)=1,n=4`
`z=i^(25)=i,n = 5,z =i^(12) = 1,......`
Thus, `z{{:(,"1,if n is odd"),(,"i, if n is even"):}`