If `alpha` is an `n^(th)` roots of unity, then `1+2alpha+3alpha^(2)+……..+nalpha^(n-1)` equals

Options

A. `n/(1-alpha)`
B. `-n/(1-alpha)`
C. `-n/(1-alpha)^(2)`
D. none of these

Answer

B

Explanation

Let `S=1+2alpha+3alpha^(2)+………………+n.alpha^(n-1)`. Then,
`alphaS=alpha+2alpha^(2)+3alpha^(3)+…………..+(n-1)alpha^(n-1)+nalpha^(n)`
`therefore S-alphaS=(1+alpha)=(alpha^(n-1))/(alpha-1) nalpha^(n)`
`rArr S(1-alpha)=(1-1)/(alpha-1)-n` `[therefore alpha^(n)=1]`
`rArr S=-n/(1-alpha)`