Question

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 1

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.