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Find the sum 1xx(2-omega)xx(2-omega^(2))...

Find the sum `1xx(2-omega)xx(2-omega^(2))+2xx(-3-omega)xx(3-omega^(2))+….+(n-1)xx(n-omega)xx(n-omega^(2))`, where `omega` is an imaginary cube root of unity.

Text Solution

Verified by Experts

The correct Answer is:
`((n(n+1))/(2))^(2) - n`
Here `T_r=(r-1)(r-omega)(r-omega)^2]=(r^3-1)`
`therefore " "S_n=underset(r=1)overset(n)Sigma(r^3-1)=[(n(n+1))/2]^2-n`
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