Question

The value of the expression

$1\cdot (2-\omega )(2-{\omega }^2)+2\cdot (3-\omega )(3-{\omega }^2)+$ _____$+(n-1)(n-\omega )(n-{\omega }^2)$

• A. $\frac{1}{4}n(n-1)(n^2+3n+4)$
• B. $n(n-1)(n^2+3n+4)$
• C. $\frac{1}{4}n(n-1)(n^2+3n-4)$
• D. None of these

Answer 1

Answer: (A)

$\frac{1}{4}n(n-1)(n^2+3n+4)$

Here, $a_n=(n-1)(n-\omega )(n-{\omega }^2)$

$\sum a_n=\sum (n-1)(n-\omega )(n-{\omega }^2)$

$=\sum (n^2-\omega n-n+\omega )(n-{\omega }^2)$

$=\sum (n^3-\omega n^2{n}^2+{\omega }^{3}n+{\omega }^{2}n-{\omega }^3)$

$=\sum (n^3-(\omega +{\omega }^3)n^2-n^2(\omega +{\omega }^2)n+n-1)[\because {\omega }^3=1]$

$=\sum (n^3-(-1)n^2-n^2+(-1)n+n-1)[\because 1+\omega +{\omega }^2=0]$

$=\sum (n^3+n^2-n^2-n+n-1)$

$=\sum (n^3-1)$

$=\sum n^3-\sum 1$

$={\left[\frac{n(n+1)}{2}\right]}^2-n$

$=\frac{n^2(n+1{)}^2}{4}-n$

$=\frac{n^4+n^2+2n^3-4n}{4}$

$=\frac{n}{4}(n^3+2n^2+n-4)$

$=\frac{n(n-1)(n^2+3n+4)}{4}$

$\therefore 1\cdot (2-\omega )(2-{\omega }^2)+2\cdot (3-\omega )(3-{\omega }^2)+$ _____$+(n-1)(n-\omega )(n-{\omega }^2)=\frac{1}{4}n(n-1)(n^2+3n+4)$