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

Question

The value of the expression 1(2-ω)(2-ω2)+(3-ω)(3-ω2)+...... +(n-1).(n-ω)(n-ω2), where, ω is an imaginary cube root of unity, is.... (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Here, Tr= (r - 1) (r -ω) (r - ω)2] =(r3-1) ∴ Sn= displaystyle ∑(r=1)n (r3-1)=[(n(n+1)/2)]2-n

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

Here, $T_{r} = (r - 1) (r - ω) (r - ω)^{2}]$
$= (r^{3} - 1)$
$∴ S_{n} = (r = 1) \sum n (r^{3} - 1) = [\frac{n ( n + 1 )}{2}]^{2} - n$

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

Here, Tr​=(r−1)(r−ω)(r−ω)2] =(r3−1) ∴Sn​=(r=1)∑n​(r3−1)=[2n(n+1)​]2−n

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

Here, $T_{r} = (r - 1) (r - ω) (r - ω)^{2}]$
$= (r^{3} - 1)$
$∴ S_{n} = (r = 1) \sum n (r^{3} - 1) = [\frac{n ( n + 1 )}{2}]^{2} - n$