If `alpha (!=1)` is a nth root of unity then `S = 1 + 3alpha+ 5alpha^2 + .......... `upto n terms is equal to

Explanation

`alpha` is the nth root of unity.
Therefore, `alpha` is root of the equation `z^(n)=1`.
Let `S=1+2alpha+3alpha^(2)+…+nalpha^(n-1)`
`rArralphaS=alpha+2alpha^(2)+3alpha^(2)+…+(n-1)alpha^(n-1)+nalpha^(n)`
On subtracting, we get
`S(1-alpha)=1+[alpha+alpha^(2)+alpha^(n-1)]-nalpha^(n)`
`rArrS(1-alpha)=1+(alpha(1-alpha^(n-1)))/(1-alpha)-nalpha^(n)`
`rArrS=1/(1-alpha)+(alpha-alpha^(n))/((1-alpha)^(2))-(nalpha^(n))/(1-alpha)`
`=1/(1-alpha)+(alpha-1)/((1-alpha)^(2))-n/(1-alpha)` `[becausealpha^(n)=1]`
`=-n/(1-alpha)`