Complex Numbers And Quadratic Equations question 163
Question: Let $ \omega $ is an imaginary cube roots of unity then the value of $ 2(\omega +1)({{\omega }^{2}}+1)+3(2\omega +1)(2{{\omega }^{2}}+1)+….. $ $ +(n+1)(n\omega +1)(n{{\omega }^{2}}+1) $ is [Orissa JEE 2002]
Options:
A) $ {{[ \frac{n(n+1)}{2} ]}^{2}}+n $
B) $ {{[ \frac{n(n+1)}{2} ]}^{2}} $
C) $ {{[ \frac{n(n+1)}{2} ]}^{2}}-n $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ 2(\omega +1)({{\omega }^{2}}+1)+3(2\omega +1)(2{{\omega }^{2}}+1)+…… $ $ +(n+1)(n\omega +1)(n{{\omega }^{2}}+1) $ = $ \sum\limits_{r=1}^{n}{(r+1)(r\omega +1)(r{{\omega }^{2}}+1)} $ = $ \sum\limits_{r=1}^{n}{(r+1)(r^{2}{{\omega }^{3}}+r\omega +r{{\omega }^{2}}+1)} $ = $ \sum\limits_{r=1}^{n}{(r+1)(r^{2}-r+1)} $ = $ \sum\limits_{r=1}^{n}{(r^{3}-r^{2}+r+r^{2}-r+1)} $ = $ \sum\limits_{r=1}^{n}{(r^{3})+\sum\limits_{r=1}^{n}{(1})} $ = $ {{[ \frac{n(n+1)}{2} ]}^{2}}+n $ .
BETA
