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

$

(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}} \\

$

Last updated date: 24th Mar 2023

•

Total views: 307.2k

•

Views today: 6.87k

Answer

Verified

307.2k+ views

Hint: In this problem we have to evaluate the given expression and it has been given that $\omega $ is the cube root of units. Let’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.

Recently Updated Pages

If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE