If `omega`is a complex nth root of unity, then `sum_(r=1)^n(ar+b)omega^(r-1)`is equal toA..`(n(n+1)a)/2`B. `(n b)/(1+n)`C. `(n a)/(omega-1)`D. none of these

If `omega`is a complex nth root of unity, then `sum_(r=1)^n(ar+b)omega

1.	If `omega`is a complex nth root of unity, then `sum_(r=1)^n(ar+b)omega^(r-1)`is equal toA..`(n(n+1)a)/2`B. `(n b)/(1+n)`C. `(n a)/(omega-1)`D. none of these
Answer» `E= sum_(r=1)^n(a+b) omega^(r-1)` `= (a+b) +((2a+b) omega + (3a+b)omega^2 + (4a+b) omega^3 + .... + (na+b)omega^(n-1)` `=b (1+ omega + omega^2 + .....+ omega^(n-1)) + a(1+ 2omega + 3omega^2 + 4 omega^3 + .... + n omega^(n-1)` S`= 1 + 2omega + 3omega^2 + 4omega^3 + ..... + nomega^(n-2)` S`omega= omega + 2omega^2 + 3omega^3 + .... + nomega^n` subtracting these equations we get `S(1-omega)= 1+ omega + omega^2 + omega^3 + .... + omega^(n-1) - n omega^n` `= 0- n omega^n = -n` `S= -n/(1-omega)` E`= 0 + aS= (-an)/(1- omega)` `E= (-na)/(1- omega) = (na)/(omega-1)` option C is corect