Q. The value of the expression 1. $(2 - \omega)\left( 2 - \omega^{2} \right) + 2(3 - \omega)\left( 3 - \omega^{2} \right) + .\ldots\ldots + .\ldots + (n - 1)(n - \omega)(n - \omega^{2})$ Where $\omega$ is an imaginary cube root of unit is
A) $(n - 1)n(n^{2} + 3n + 4)/4$
B) $(n - 1)n(n^{2} + 3n + 4)/2$
C) $(n + 1)n(n^{2} + 3n + 4)/2$
D) None of the above
Explanation
(a)
$T_{r} = r\left\lbrack (r + 1) - \omega \right\rbrack\lbrack(r + 1) - \omega^{2}\rbrack$
$= r\lbrack(r + 1)^{2} - \left( \omega + \omega^{2} \right)(r + 1) + \omega^{3}\rbrack$
$= r\left\lbrack (r + 1)^{2} - ( - 1)(r + 1) + 1 \right\rbrack = r^{3} + 3r^{2} + 3r$
$\therefore\ \ \ \ \ \ \ \sum_{r = 1}^{n - 1}{T_{r} = \sum_{r = 1}^{n - 1}{(r^{3} + 3r^{2} + 3r)}}$
$= \frac{1}{4}(n - 1)^{2}(n)^{2} + 3.\frac{1}{6}(n - 1)n(2n - 1) + 3.\frac{1}{2}(n - 1)n\$
$= \frac{1}{4}\ (n - 1)n(n^{2} + 3n + 4)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.