If `omega` is a complex cube root of unity, then the value of the expression `1(2-omega)(2-omega^2)+2(3-omega)(3-omega^2) +...+(n-1) (n-omega)(n-omega^2) (n>=2) ` is equal to (A) `(n^2(n+1)^2)/4 - n` (B) `(n^2(n+1)^2)/4 +n` (C) `(n^2(n+1))/4 -n` (D) `(n(n+1)^2)/4 -n`

If `omega` is a complex cube root of unity, then the value of the expr

1.	If `omega` is a complex cube root of unity, then the value of the expression `1(2-omega)(2-omega^2)+2(3-omega)(3-omega^2) +...+(n-1) (n-omega)(n-omega^2) (n>=2) ` is equal to (A) `(n^2(n+1)^2)/4 - n` (B) `(n^2(n+1)^2)/4 +n` (C) `(n^2(n+1))/4 -n` (D) `(n(n+1)^2)/4 -n`
Answer» We can write the given expression as , `S = sum_(m=1)^n(m-1)(m-omega)(m-omega^2)` So , general term in this expression can be given as, `T_m = (m-1)(m-omega)(m-omega^2)` `= (m-1)(m^2-(omega+omega^2)m+omega^3)` As, `1+omega+omega^2 = 0=>omega+omega^2 = -1` `:. T_m = (m-1)(m^2+m+1)` (As `omega^3 = 1`) We know, `(a-1)(a^2+a+1) = a^3-1``:. T_m=m^3-1` `:. S = sum_(m=1)^n (m^3-1) = sum_(m=1)^n m^3- sum_(m=1)^n 1` `=> S= ((n(n+1))/2)^2+n = 1/4(n^2)(n+1)^2-n` So, option `A` is the correct option.