If ω is an imaginary cube root of unity, then the value of (2-ω)(2-ω2)+2(3-ω)(3-ω2)+.....+(n-1)(n-ω)(n-ω2)

Question

If ω is an imaginary cube root of unity, then the value of (2-ω)(2-ω2)+2(3-ω)(3-ω2)+.....+(n-1)(n-ω)(n-ω2) (A) (n2/4)(n+1)2-n (B) (n2/4)(n+1)2+n (C) (n2/4)(n+1)2 (D) (n2/4)(n+1)-n

Tardigrade Admin · Accepted Answer

displaystyle ∑r=2n(r-1)(r-ω2)= displaystyle ∑r=2n(r3-1)=[(n2(n+1)2/4)-1]-(n-1)=(n2(n+1)2/4)-n

Q. If $ω$ is an imaginary cube root of unity, then the value of $(2 - ω) (2 - ω^{2}) + 2 (3 - ω) (3 - ω^{2}) + ..... + (n - 1) (n - ω) (n - ω^{2})$

$\frac{n ^{2}}{4} (n + 1)^{2} - n$

$\frac{n ^{2}}{4} (n + 1)^{2} + n$

$\frac{n ^{2}}{4} (n + 1)^{2}$

$\frac{n ^{2}}{4} (n + 1) - n$

$r = 2 \sum n (r - 1) (r - ω^{2}) = r = 2 \sum n (r^{3} - 1) =$ $[\frac{n ^{2} ( n + 1 ) ^{2}}{4} - 1] - (n - 1) = \frac{n ^{2} ( n + 1 ) ^{2}}{4} - n$

Q. If ω is an imaginary cube root of unity, then the value of (2−ω)(2−ω2)+2(3−ω)(3−ω2)+.....+(n−1)(n−ω)(n−ω2)

4n2​(n+1)2−n

4n2​(n+1)2+n

4n2​(n+1)2

4n2​(n+1)−n