Question

The value of sum `Sigma_(n=1)^(13) (i^n + i^(n+1))` where `i= sqrt(-1)` equals
A. i
B. i-1
C. `-i`
D. 0

Answer 1

Correct Answer - D
`underset(n=1)overset(13)Sigma(i^n+i^(n+1))=underset(n=1)overset(13)Sigmai^n(1+i)=(1+i)underset(n=1)overset(13)Sigma i^n`
`=(1+i)(i+i^2+i^3.....+i^(13))=(1+i)[(i-(1-i^(13)))/(1-i)] `
`=(l+i)[(i(1-i))/(1-i)]=(1+i)=i-1`
Alternate Solution
Since sum of any four comsecutive powers fo iota is zero
` therefore underset(n=1) overset(13)Sigma (i^n+i^(n+1)+....+i^(13))`
`+(i^2+i^3 +.....+i^(14))=i+i^2=i-1`