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(D) Given that $\omega$ is a complex cube root of unity,we have $1+\omega+\omega^2 = 0$ and $\omega^3 = 1$. Consider the expression $r(r+1-\omega)(r+1

Vedclass · Accepted Answer

$3015$

Vedclass · Answer

(D) Given that $\omega$ is a complex cube root of unity,we have $1+\omega+\omega^2 = 0$ and $\omega^3 = 1$. Consider the expression $r(r+1-\omega)(r+1-\omega^2)$. Expanding this,we get $r[(r+1)^2 - (r+1)(\omega+\omega^2) + \omega^3]$. Since $\omega+\omega^2 = -1$ and $\omega^3 = 1$,the expression becomes $r[(r+1)^2 + (r+1) + 1] = r(r^2+2r+1+r+1+1) = r(r^2+3r+3) = r^3+3r^2+3r$. Now,we calculate the sum $\sum_{r=1}^9 (r^3+3r^2+3r) = \sum_{r=1}^9 r^3 + 3\sum_{r=1}^9 r^2 + 3\sum_{r=1}^9 r$. Using the standard summation formulas: $\sum_{r=1}^9 r^3 = [\frac{9(10)}{2}]^2 = 45^2 = 2025$. $3\sum_{r=1}^9 r^2 = 3 	imes \frac{9(10)(19)}{6} = 3 	imes 285 = 855$. $3\sum_{r=1}^9 r = 3 	imes \frac{9(10)}{2} = 3 	imes 45 = 135$. Adding these values: $2025 + 855 + 135 = 3015$. Thus,the correct option is $D$.

If $\omega$ is a complex cube root of unity,then $\sum_{r=1}^9 r(r+1-\omega)(r+1-\omega^2) = $

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