Question

# The value of the expression $1\left( {2 - \omega } \right)\left( {2 - {\omega ^2}} \right) + 2\left( {3 - \omega } \right) \times \left( {3 - {\omega ^2}} \right) + .......... + \left( {n - 1} \right)\left( {n - \omega } \right)\left( {n - {\omega ^2}} \right),$ where ω is an imaginary cube root of unity is

Hint: At first we’ll find the general term of the given expression. Then we’ll write it as
${S_n} = \sum\limits_{r = 1}^n {{T_r}}$, where ${T_r} = r(r + 1 - \omega )(r + 1 - {\omega ^2})$ , then also we should always check the number of terms the expression is containing as in the given expression we have (n-1) terms, then by simplifying further we get our answer.

Given data: the expression $1\left( {2 - \omega } \right)\left( {2 - {\omega ^2}} \right) + 2\left( {3 - \omega } \right) \times \left( {3 - {\omega ^2}} \right) + .......... + \left( {n - 1} \right)\left( {n - \omega } \right)\left( {n - {\omega ^2}} \right)$
From the given expression we can say that,
${r^{th}}$ term or ${T_r} = r(r + 1 - \omega )(r + 1 - {\omega ^2})$
On expanding the right-hand side
$= r({r^2} + r - r{\omega ^2} + r + 1 - {\omega ^2} - r\omega - \omega + {\omega ^3})$
$= r\left[ {{r^2} + r(1 - {\omega ^2} - \omega + 1) + 1 - {\omega ^2} - \omega + {\omega ^3}} \right]$
We know that the sum of the cube root of unity is zero
i.e.$1 + \omega + {\omega ^2} = 0$
$\Rightarrow 1 = - \omega - {\omega ^2}$
And ${\omega ^3} = 1$
$\therefore r\left[ {{r^2} + r(1 - {\omega ^2} - \omega + 1) + 1 - {\omega ^2} - \omega + {\omega ^3}} \right] = r\left[ {{r^2} + r(1 + 1 + 1) + 1 + 1 + 1} \right]$
$= r\left[ {{r^2} + 3r + 3} \right]$
Now, opening the brackets
$r\left[ {{r^2} + 3r + 3} \right] = {r^3} + 3{r^2} + 3r$
Now we can say that,
$1\left( {2 - \omega } \right)\left( {2 - {\omega ^2}} \right) + 2\left( {3 - \omega } \right) \times \left( {3 - {\omega ^2}} \right) + .......... + \left( {n - 1} \right)\left( {n - \omega } \right)\left( {n - {\omega ^2}} \right) = \sum\limits_{r = 1}^{n - 1} {{r^3}} + 3{r^2} + 3r$
It is well known that,
$\sum\limits_{r = 1}^n {(A + B + C} ) = \sum\limits_{r = 1}^n A + \sum\limits_{r = 1}^n B + \sum\limits_{r = 1}^n C$
$\therefore \sum\limits_{r = 1}^n {{r^3}} + 3{r^2} + 3r = \sum\limits_{r = 1}^n {{r^3}} + 3\sum\limits_{r = 1}^n {{r^2}} + 3\sum\limits_{r = 1}^n r$
Now, as we all know
$\sum\limits_{r = 1}^n r = \dfrac{{n\left( {n + 1} \right)}}{2} \\ \sum\limits_{r = 1}^n {{r^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} \\ \sum\limits_{r = 1}^n {{r^3}} = {\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2} \\$
$\therefore \sum\limits_{r = 1}^{n - 1} {{r^3}} + 3\sum\limits_{r = 1}^{n - 1} {{r^2}} + 3\sum\limits_{r = 1}^{n - 1} r = {\left( {\dfrac{{(n - 1)n}}{2}} \right)^2} + 3\dfrac{{(n - 1)n\left( {2(n - 1) + 1} \right)}}{6} + 3\dfrac{{(n - 1)n}}{2}$
Now, on solving the left-hand side
i.e.${\left( {\dfrac{{(n - 1)n}}{2}} \right)^2} + 3\dfrac{{(n - 1)n\left( {2(n - 1) + 1} \right)}}{6} + 3\dfrac{{(n - 1)n}}{2} = \dfrac{{{{(n - 1)}^2}{n^2}}}{4} + \dfrac{{(n - 1)n\left( {2n - 1} \right)}}{2} + \dfrac{{3(n - 1)n}}{2}$
using ${(a - b)^2} = {a^2} + {b^2} - 2ab$ and simplifying the brackets,
$= \dfrac{{\left( {{n^2} + 1 - 2n} \right){n^2}}}{4} + \dfrac{{n\left( {2{n^2} - n - 2n + 1} \right)}}{2} + \dfrac{{3{n^2} - 3n}}{2}$
Again simplifying the brackets further, we get,
$= \dfrac{{{n^4} + {n^2} - 2{n^3}}}{4} + \dfrac{{2{n^3} - {n^2} - 2{n^2} + n}}{2} + \dfrac{{3{n^2} - 3n}}{2}$
$= \dfrac{{{n^4} + {n^2} - 2{n^3}}}{4} + \dfrac{{2{n^3} - {n^2} - 2{n^2} + n + 3{n^2} - 3n}}{2}$
$= \dfrac{{{n^4} + {n^2} - 2{n^3}}}{4} + \dfrac{{2{n^3} - 2n}}{2}$
Now, adding both the terms by taking LCM
$= \dfrac{{{n^4} + {n^2} - 2{n^3} + 4{n^3} - 4n}}{4}$
$= \dfrac{{{n^4} + 2{n^3} + {n^2} - 4n}}{4}$
$= \dfrac{n}{4}\left( {{n^3} + 2{n^2} + n - 4} \right)$
$\therefore 1\left( {2 - \omega } \right)\left( {2 - {\omega ^2}} \right) + 2\left( {3 - \omega } \right) \times \left( {3 - {\omega ^2}} \right) + .......... + \left( {n - 1} \right)\left( {n - \omega } \right)\left( {n - {\omega ^2}} \right) = \dfrac{n}{4}\left( {{n^3} + 2{n^2} + n - 4} \right)$
Note: In this question, we have given the cube root of unity. There are a total 3 cube root of unity namely $1,\omega ,{\omega ^2}$
Where $\omega ,{\omega ^2}$are the imaginary roots having value as
$\omega = - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}$ , and
${\omega ^2} = - \dfrac{1}{2} - i\dfrac{{\sqrt 3 }}{2}$
From the values of $\omega ,{\omega ^2}$, we can say that sum of cube roots of unity is zero
i.e. $1 + \omega + {\omega ^2} = 1 - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2} - \dfrac{1}{2} - i\dfrac{{\sqrt 3 }}{2}$
$= 0$