Q. The value of the expression 1 . $(2 - \omega)\left( 2 - \omega^{2} \right) + 2.(3 - \omega)\left( 3 - \omega^{2} \right) + ...\ldots + (n - 1).(n - \omega)\left( n - \omega^{2} \right)$ Where $\omega$ is an imaginary cube root of unity, is
A) $\frac{1}{2}(n - 1)\ n\ \left( n^{2} + 3n + 4 \right)$
B) $\frac{1}{4}(n - 1)\ n\ \left( n^{2} + 3n + 4 \right)$
C) $\frac{1}{2}(n + 1)\ n\ \left( n^{2} + 3n + 4 \right)$
D) $\frac{1}{4}(n + 1)\ n\ \left( n^{2} + 3n + 4 \right)$
Explanation
(b)
$r$th term of the given series
$= r\left\lbrack (r + 1) - \omega \right\rbrack\left\lbrack (r + 1) - \omega^{2} \right\rbrack$
$= r\lbrack(r + 1)^{2} - \left( \omega + \omega^{2} \right)(r + 1) + \omega^{3}\rbrack$
$= r\lbrack(r + 1)^{2} - ( - 1)(r + 1) + 1\rbrack$
$= r\left( r^{2} + 3r + 3 \right) = r^{3} + 3r^{2} + 3r$
Thus, sum of the give series
$= \ \sum_{r = 1}^{(n - 1)}{(r^{3} + 3r^{2} + 3r)}$
$= \frac{1}{4}(n - 1)^{2}n^{2} + 3.\frac{1}{6}(n - 1)(n)(2n - 1) + 3.\frac{1}{2}(n - 1)n$
$= \frac{1}{4}(n - 1)n\ (n^{2} + 3n + 4)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.