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Question

Answers

$

(a){\text{ }}{\left\{ {\dfrac{{n(n + 1)}}{2}} \right\}^2} \\

(a){\text{ }}{\left\{ {\dfrac{{n(n + 1)}}{2}} \right\}^2} - {\text{n}} \\

(a){\text{ }}{\left\{ {\dfrac{{n(n + 1)}}{2}} \right\}^2} + {\text{n}} \\

(a){\text{ None of the above}} \\

$

Answer
Verified

Hint: In this problem we have to evaluate the given expression and it has been given that $\omega $ is the cube root of units. Letâ€™s understand what does a cube root of unity means the solution of ${\left( 1 \right)^{\dfrac{1}{3}}} = \left( {1,\omega ,{\omega ^2}} \right)$, thus $\omega $ is known as cube root of unit. Use the various properties of cube root of unity like $\left( {1 + \omega + {\omega ^2} = 0} \right){\text{ & }}\left( {{\omega ^3} = 1} \right)$ along with the formula for sum of squares of first n natural numbers to get the answer.

Complete step-by-step answer:

Given expression is

$1.\left( {2 - \omega } \right)\left( {2 - {\omega ^2}} \right) + 2.\left( {3 - \omega } \right)\left( {3 - {\omega ^2}} \right) + ............. + \left( {n - 1} \right)\left( {n - \omega } \right)\left( {n - {\omega ^2}} \right)$

Now it is given that $\omega $ is an imaginary cube root of unity

Therefore $\left( {1 + \omega + {\omega ^2} = 0} \right){\text{ & }}\left( {{\omega ^3} = 1} \right)$

Now the first term of the series is

$1.\left( {2 - \omega } \right)\left( {2 - {\omega ^2}} \right)$

Now simplify it we have,

$1.\left( {2 - \omega } \right)\left( {2 - {\omega ^2}} \right) = 4 - 2{\omega ^2} - 2\omega + {\omega ^3} = 4 - 2\left( {\omega + {\omega ^2}} \right) + {\omega ^3}$

Now substitute the value of $\left( {\omega + {\omega ^2}} \right)$ which is -1 and the value of ${\omega ^3}$ which is 1.

$ \Rightarrow 1.\left( {2 - \omega } \right)\left( {2 - {\omega ^2}} \right) = 4 - 2\left( { - 1} \right) + 1 = 7$

$ \Rightarrow 1.\left( {2 - \omega } \right)\left( {2 - {\omega ^2}} \right) = {2^3} - 1$

Now the second term of the series is $2.\left( {3 - \omega } \right)\left( {3 - {\omega ^2}} \right)$

Now simplify it we have,

$ \Rightarrow 2.\left( {3 - \omega } \right)\left( {3 - {\omega ^2}} \right) = 2\left( {9 - 3{\omega ^2} - 3\omega + {\omega ^3}} \right) = 2\left( {9 - 3\left( { - 1} \right) + 1} \right) = 2\left( {13} \right) = 26$

$ \Rightarrow 2.\left( {3 - \omega } \right)\left( {3 - {\omega ^2}} \right) = {3^3} - 1$

Similarly ${n^{th}}$ term of the series is $\left( {{n^3} - 1} \right)$

So, the series converted into

$\left( {{2^3} - 1} \right) + \left( {{3^3} - 1} \right) + .................. + \left( {{n^3} - 1} \right)$

Now the sum of this series is

${S_n} = \sum\limits_{r = 1}^n {\left( {{r^3} - 1} \right)} $

$ \Rightarrow {S_n} = \sum\limits_{r = 1}^n {{r^3} - } \sum\limits_{r = 1}^n 1 $

Now as we know summation of $\sum\limits_{r = 1}^n {{r^3}} = {\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2}$ and $\sum\limits_{r = 1}^n 1 = n$

So substitute these values in above equation we have,

$ \Rightarrow {S_n} = {\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2} - n$

So this is the required answer.

Hence, option (b) is correct.

Note: Whenever we face such types of problems involving cube root of unity the key point is to have a good grasp over the formula involving cube root of unity, some of which are mentioned above. The basic understanding about the definition of the cube root of unity along with formula implementation after simplification will help you get on the right track to reach the answer.

Complete step-by-step answer:

Given expression is

$1.\left( {2 - \omega } \right)\left( {2 - {\omega ^2}} \right) + 2.\left( {3 - \omega } \right)\left( {3 - {\omega ^2}} \right) + ............. + \left( {n - 1} \right)\left( {n - \omega } \right)\left( {n - {\omega ^2}} \right)$

Now it is given that $\omega $ is an imaginary cube root of unity

Therefore $\left( {1 + \omega + {\omega ^2} = 0} \right){\text{ & }}\left( {{\omega ^3} = 1} \right)$

Now the first term of the series is

$1.\left( {2 - \omega } \right)\left( {2 - {\omega ^2}} \right)$

Now simplify it we have,

$1.\left( {2 - \omega } \right)\left( {2 - {\omega ^2}} \right) = 4 - 2{\omega ^2} - 2\omega + {\omega ^3} = 4 - 2\left( {\omega + {\omega ^2}} \right) + {\omega ^3}$

Now substitute the value of $\left( {\omega + {\omega ^2}} \right)$ which is -1 and the value of ${\omega ^3}$ which is 1.

$ \Rightarrow 1.\left( {2 - \omega } \right)\left( {2 - {\omega ^2}} \right) = 4 - 2\left( { - 1} \right) + 1 = 7$

$ \Rightarrow 1.\left( {2 - \omega } \right)\left( {2 - {\omega ^2}} \right) = {2^3} - 1$

Now the second term of the series is $2.\left( {3 - \omega } \right)\left( {3 - {\omega ^2}} \right)$

Now simplify it we have,

$ \Rightarrow 2.\left( {3 - \omega } \right)\left( {3 - {\omega ^2}} \right) = 2\left( {9 - 3{\omega ^2} - 3\omega + {\omega ^3}} \right) = 2\left( {9 - 3\left( { - 1} \right) + 1} \right) = 2\left( {13} \right) = 26$

$ \Rightarrow 2.\left( {3 - \omega } \right)\left( {3 - {\omega ^2}} \right) = {3^3} - 1$

Similarly ${n^{th}}$ term of the series is $\left( {{n^3} - 1} \right)$

So, the series converted into

$\left( {{2^3} - 1} \right) + \left( {{3^3} - 1} \right) + .................. + \left( {{n^3} - 1} \right)$

Now the sum of this series is

${S_n} = \sum\limits_{r = 1}^n {\left( {{r^3} - 1} \right)} $

$ \Rightarrow {S_n} = \sum\limits_{r = 1}^n {{r^3} - } \sum\limits_{r = 1}^n 1 $

Now as we know summation of $\sum\limits_{r = 1}^n {{r^3}} = {\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2}$ and $\sum\limits_{r = 1}^n 1 = n$

So substitute these values in above equation we have,

$ \Rightarrow {S_n} = {\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2} - n$

So this is the required answer.

Hence, option (b) is correct.

Note: Whenever we face such types of problems involving cube root of unity the key point is to have a good grasp over the formula involving cube root of unity, some of which are mentioned above. The basic understanding about the definition of the cube root of unity along with formula implementation after simplification will help you get on the right track to reach the answer.

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