Answer
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Hint: First of all, write the ${m^{th}}$ term of the given series. And then, use the property of $\omega $ to simplify the given condition. Next, add and subtract $m$ to the expression and use the property $1 + \omega + {\omega ^2} = 0$ to simplify the ${m^{th}}$ term of given expression. Hence, use the direct formula of summation to find the required answer.
Complete step by step answer:
The ${m^{th}}$ term of the given series $1\left( {2 - \omega } \right)\left( {2 - {\omega ^2}} \right) + 2\left( {3 - \omega } \right)\left( {3 - {\omega ^2}} \right) + ... + \left( {n - 1} \right)\left( {n - \omega } \right)\left( {n - {\omega ^2}} \right)$ can be written as $\left( {m - 1} \right)\left( {m - \omega } \right)\left( {m - {\omega ^2}} \right)$.
We can simplify the expression for the ${m^{th}}$ term of the series.
$\left( {m - 1} \right)\left( {{m^2} - m{\omega ^2} - m\omega + {\omega ^3}} \right)$
It is known that ${\omega ^3} = 1$ since $\omega $ is the cube root of the unity.
Therefore the expression $\left( {m - 1} \right)\left( {{m^2} - m{\omega ^2} - m\omega + {\omega ^3}} \right)$ becomes
$\left( {m - 1} \right)\left( {{m^2} - m{\omega ^2} - m\omega + 1} \right)$
Adding $m$and subtracting $m$ in the second factor of the expression and simplifying, we get
$
= \left( {m - 1} \right)\left( {{m^2} - m{\omega ^2} - m\omega - m + m + 1} \right) \\
= \left( {m - 1} \right)\left( {{m^2} - \left( {{\omega ^2} + \omega + 1} \right)m + m + 1} \right) \\
$
It is known, the property of $\omega $, that is $1 + \omega + {\omega ^2} = 0$. Thus the expression $\left( {m - 1} \right)\left( {{m^2} - \left( {{\omega ^2} + \omega + 1} \right)m + m + 1} \right)$ reduces to
$ = \left( {m - 1} \right)\left( {{m^2} + m + 1} \right)$
Simplifying the expression for the ${m^{th}}$ term of the series, we get
$
= \left( {m - 1} \right)\left( {{m^2} + m + 1} \right) \\
= {m^3} + {m^2} + m - {m^2} - m - 1 \\
= {m^3} - 1 \\
$
The sum of the given series $1\left( {2 - \omega } \right)\left( {2 - {\omega ^2}} \right) + 2\left( {3 - \omega } \right)\left( {3 - {\omega ^2}} \right) + ... + \left( {n - 1} \right)\left( {n - \omega } \right)\left( {n - {\omega ^2}} \right)$ can be evaluated by taking the summation of the ${m^{th}}$ term of the series from $m = 1$ to $m = n$.
Thus the sum of given series is
$\sum\limits_{m = 1}^n {{m^3} - 1} $
Solving the expression $\sum\limits_{m = 1}^n {{m^3} - 1} $, we get
$
= \sum\limits_{m = 1}^n {{m^3}} - \sum\limits_{m = 1}^n 1 \\
= {\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2} - n \\
$
The sum of the given series $1\left( {2 - \omega } \right)\left( {2 - {\omega ^2}} \right) + 2\left( {3 - \omega } \right)\left( {3 - {\omega ^2}} \right) + ... + \left( {n - 1} \right)\left( {n - \omega } \right)\left( {n - {\omega ^2}} \right)$ is ${\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2} - n$.
Thus the option B is the correct answer.
Note: The summation of ${x^3}$ over the range 1 to $n$ is ${\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2}$. The summation of 1 for $n$times is equal to $n$. Similarly, The summation of ${x^2}$ over the range 1 to $n$ is $\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$. And, summation of $x$ over the range 1 to $n$ is $\dfrac{{n\left( {n + 1} \right)}}{2}$.
Complete step by step answer:
The ${m^{th}}$ term of the given series $1\left( {2 - \omega } \right)\left( {2 - {\omega ^2}} \right) + 2\left( {3 - \omega } \right)\left( {3 - {\omega ^2}} \right) + ... + \left( {n - 1} \right)\left( {n - \omega } \right)\left( {n - {\omega ^2}} \right)$ can be written as $\left( {m - 1} \right)\left( {m - \omega } \right)\left( {m - {\omega ^2}} \right)$.
We can simplify the expression for the ${m^{th}}$ term of the series.
$\left( {m - 1} \right)\left( {{m^2} - m{\omega ^2} - m\omega + {\omega ^3}} \right)$
It is known that ${\omega ^3} = 1$ since $\omega $ is the cube root of the unity.
Therefore the expression $\left( {m - 1} \right)\left( {{m^2} - m{\omega ^2} - m\omega + {\omega ^3}} \right)$ becomes
$\left( {m - 1} \right)\left( {{m^2} - m{\omega ^2} - m\omega + 1} \right)$
Adding $m$and subtracting $m$ in the second factor of the expression and simplifying, we get
$
= \left( {m - 1} \right)\left( {{m^2} - m{\omega ^2} - m\omega - m + m + 1} \right) \\
= \left( {m - 1} \right)\left( {{m^2} - \left( {{\omega ^2} + \omega + 1} \right)m + m + 1} \right) \\
$
It is known, the property of $\omega $, that is $1 + \omega + {\omega ^2} = 0$. Thus the expression $\left( {m - 1} \right)\left( {{m^2} - \left( {{\omega ^2} + \omega + 1} \right)m + m + 1} \right)$ reduces to
$ = \left( {m - 1} \right)\left( {{m^2} + m + 1} \right)$
Simplifying the expression for the ${m^{th}}$ term of the series, we get
$
= \left( {m - 1} \right)\left( {{m^2} + m + 1} \right) \\
= {m^3} + {m^2} + m - {m^2} - m - 1 \\
= {m^3} - 1 \\
$
The sum of the given series $1\left( {2 - \omega } \right)\left( {2 - {\omega ^2}} \right) + 2\left( {3 - \omega } \right)\left( {3 - {\omega ^2}} \right) + ... + \left( {n - 1} \right)\left( {n - \omega } \right)\left( {n - {\omega ^2}} \right)$ can be evaluated by taking the summation of the ${m^{th}}$ term of the series from $m = 1$ to $m = n$.
Thus the sum of given series is
$\sum\limits_{m = 1}^n {{m^3} - 1} $
Solving the expression $\sum\limits_{m = 1}^n {{m^3} - 1} $, we get
$
= \sum\limits_{m = 1}^n {{m^3}} - \sum\limits_{m = 1}^n 1 \\
= {\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2} - n \\
$
The sum of the given series $1\left( {2 - \omega } \right)\left( {2 - {\omega ^2}} \right) + 2\left( {3 - \omega } \right)\left( {3 - {\omega ^2}} \right) + ... + \left( {n - 1} \right)\left( {n - \omega } \right)\left( {n - {\omega ^2}} \right)$ is ${\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2} - n$.
Thus the option B is the correct answer.
Note: The summation of ${x^3}$ over the range 1 to $n$ is ${\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2}$. The summation of 1 for $n$times is equal to $n$. Similarly, The summation of ${x^2}$ over the range 1 to $n$ is $\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$. And, summation of $x$ over the range 1 to $n$ is $\dfrac{{n\left( {n + 1} \right)}}{2}$.
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