If `omega` is a complex nth root of unity, then `underset(r=1)overset(n)(ar+b)omega^(r-1)` is equal to

Options

A. `(n(n+1))a)/(a)`
B. `(nb)/(1-n)`
C. `(na)/(omega-1)`
D. none of these

Answer

C

Explanation

`S=sum_(r=1)^(n)(ar+b)omega^(r-1)`
`=(a+b)+(2a+b)omega+(3a+b)omega^(2)+..+(na+b)omega^(n-1)`
`=b(1+omega+omega^(2)+..+omega^(n-1))+a(1+2omega+3omega^(2)+…+nomega^(n-1))`
`=0+a(1+2omega+2omega^(2)+..+nomega^(n-1))`
Now, `S_(1)=1+2omega+3omega^(2)+..+nomega^(n-1)`
`thereforeomegaS_(1)=omega+2omega^(2)+...+(n-1)omega^(n-1)+nomega^(n)`
Subtracting, we get
`S_(1)(1-omega)=1+omega+omega^(2)+..+omega^(n-1)-nomega^(n)=-n`
`(asomega^(n)=1)`
`rArrS_(1)=n/(omega-1)`
`thereforeS=(na)/(omega-1)`