Solveeit Logo
Login

Question

Question: The value of the expression \(1\left( {2 - \omega } \right)\left( {2 - {\omega ^2}} \right) + 2\left...

The value of the expression 1(2−ω)(2−ω2)+2(3−ω)×(3−ω2)+..........+(n−1)(n−ω)(n−ω2),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

Explanation

Solution

At first we’ll find the general term of the given expression. Then we’ll write it as
Sn=∑r=1nTr{S_n} = \sum\limits_{r = 1}^n {{T_r}} , where Tr=r(r+1−ω)(r+1−ω2){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.

Complete step-by-step answer:
Given data: the expression 1(2−ω)(2−ω2)+2(3−ω)×(3−ω2)+..........+(n−1)(n−ω)(n−ω2)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,
rth{r^{th}} term or Tr=r(r+1−ω)(r+1−ω2){T_r} = r(r + 1 - \omega )(r + 1 - {\omega ^2})
On expanding the right-hand side
=r(r2+r−rω2+r+1−ω2−rω−ω+ω3)= r({r^2} + r - r{\omega ^2} + r + 1 - {\omega ^2} - r\omega - \omega + {\omega ^3})
=r[r2+r(1−ω2−ω+1)+1−ω2−ω+ω3]= 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+ω+ω2=01 + \omega + {\omega ^2} = 0
⇒1=−ω−ω2\Rightarrow 1 = - \omega - {\omega ^2}
And ω3=1{\omega ^3} = 1
∴r[r2+r(1−ω2−ω+1)+1−ω2−ω+ω3]=r[r2+r(1+1+1)+1+1+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[r2+3r+3]= r\left[ {{r^2} + 3r + 3} \right]
Now, opening the brackets
r[r2+3r+3]=r3+3r2+3rr\left[ {{r^2} + 3r + 3} \right] = {r^3} + 3{r^2} + 3r
Now we can say that,
1(2−ω)(2−ω2)+2(3−ω)×(3−ω2)+..........+(n−1)(n−ω)(n−ω2)=∑r=1n−1r3+3r2+3r1\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,
∑r=1n(A+B+C)=∑r=1nA+∑r=1nB+∑r=1nC\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
∴∑r=1nr3+3r2+3r=∑r=1nr3+3∑r=1nr2+3∑r=1nr\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

∑r=1nr=n(n+1)2 ∑r=1nr2=n(n+1)(2n+1)6 ∑r=1nr3=(n(n+1)2)2  \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} \\\

∴∑r=1n−1r3+3∑r=1n−1r2+3∑r=1n−1r=((n−1)n2)2+3(n−1)n(2(n−1)+1)6+3(n−1)n2\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.((n−1)n2)2+3(n−1)n(2(n−1)+1)6+3(n−1)n2=(n−1)2n24+(n−1)n(2n−1)2+3(n−1)n2{\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=a2+b2−2ab{(a - b)^2} = {a^2} + {b^2} - 2ab and simplifying the brackets,
=(n2+1−2n)n24+n(2n2−n−2n+1)2+3n2−3n2= \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,
=n4+n2−2n34+2n3−n2−2n2+n2+3n2−3n2= \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}
=n4+n2−2n34+2n3−n2−2n2+n+3n2−3n2= \dfrac{{{n^4} + {n^2} - 2{n^3}}}{4} + \dfrac{{2{n^3} - {n^2} - 2{n^2} + n + 3{n^2} - 3n}}{2}
Now adding the like terms,
=n4+n2−2n34+2n3−2n2= \dfrac{{{n^4} + {n^2} - 2{n^3}}}{4} + \dfrac{{2{n^3} - 2n}}{2}
Now, adding both the terms by taking LCM
=n4+n2−2n3+4n3−4n4= \dfrac{{{n^4} + {n^2} - 2{n^3} + 4{n^3} - 4n}}{4}
Now adding the like terms
=n4+2n3+n2−4n4= \dfrac{{{n^4} + 2{n^3} + {n^2} - 4n}}{4}
Taking n common from every term
=n4(n3+2n2+n−4)= \dfrac{n}{4}\left( {{n^3} + 2{n^2} + n - 4} \right)
∴1(2−ω)(2−ω2)+2(3−ω)×(3−ω2)+..........+(n−1)(n−ω)(n−ω2)=n4(n3+2n2+n−4)\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,ω,ω21,\omega ,{\omega ^2}
Where ω,ω2\omega ,{\omega ^2}are the imaginary roots having value as
ω=−12+i32\omega = - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2} , and
ω2=−12−i32{\omega ^2} = - \dfrac{1}{2} - i\dfrac{{\sqrt 3 }}{2}
From the values of ω,ω2\omega ,{\omega ^2}, we can say that sum of cube roots of unity is zero
i.e. 1+ω+ω2=1−12+i32−12−i321 + \omega + {\omega ^2} = 1 - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2} - \dfrac{1}{2} - i\dfrac{{\sqrt 3 }}{2}
=0= 0