Solveeit Logo
Login

Question

Question: If \(\omega \) is an imaginary cube root of unity, then the value of \(1\left( 2-\omega \right)\left...

If ω\omega is an imaginary cube root of unity, then the value of 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)\left( 3-{{\omega }^{2}} \right)+....+\left( n-1 \right)\left( n-\omega \right)\left( n-{{\omega }^{2}} \right) is?
(a) n(n+1)2−n\dfrac{n\left( n+1 \right)}{2}-n
(b) n2(n+1)24−n\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}-n
(c) n(n+1)2+n\dfrac{n\left( n+1 \right)}{2}+n
(d) n2(n+1)24+n\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}+n

Explanation

Solution

Assume the given expression as E. Multiply the terms present in the expression to form a general pattern. Use the formulas 1+ω+ω2=01+\omega +{{\omega }^{2}}=0 and ω3=1{{\omega }^{3}}=1 to simplify the expression. Now, form the summation series by cancelling the common terms and use the formulas ∑1nn3=(n(n+1)2)2\sum\limits_{1}^{n}{{{n}^{3}}}={{\left( \dfrac{n\left( n+1 \right)}{2} \right)}^{2}} and ∑1n1=n\sum\limits_{1}^{n}{1}=n to get the answer. Use the algebraic identity (a+b)(a−b)=a2−b2\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}} for the simplification.

Complete step-by-step solution:
Here we have been provided with the expression1(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)\left( 3-{{\omega }^{2}} \right)+....+\left( n-1 \right)\left( n-\omega \right)\left( n-{{\omega }^{2}} \right), where ω\omega is an imaginary cube root of unity and we are asked to find its value. Let us assume the expression as E, so we have,
⇒E=1(2−ω)(2−ω2)+2(3−ω)(3−ω2)+....+(n−1)(n−ω)(n−ω2)\Rightarrow E=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, multiplying the expression in each term we get,
⇒E=1(22−2(ω+ω2)+ω3)+2(32−3(ω+ω2)+ω3)+....+(n−1)(n2−n(ω+ω2)+ω3)\Rightarrow E=1\left( {{2}^{2}}-2\left( \omega +{{\omega }^{2}} \right)+{{\omega }^{3}} \right)+2\left( {{3}^{2}}-3\left( \omega +{{\omega }^{2}} \right)+{{\omega }^{3}} \right)+....+\left( n-1 \right)\left( {{n}^{2}}-n\left( \omega +{{\omega }^{2}} \right)+{{\omega }^{3}} \right)
We know that 1+ω+ω2=01+\omega +{{\omega }^{2}}=0, so we get ω+ω2=−1\omega +{{\omega }^{2}}=-1. Also, multiplying both the sides of the expression 1+ω+ω2=01+\omega +{{\omega }^{2}}=0 with ω\omega we get,
⇒ω+ω2+ω3=0 ⇒ω3=−(ω+ω2) ⇒ω3=1 \begin{aligned} & \Rightarrow \omega +{{\omega }^{2}}+{{\omega }^{3}}=0 \\\ & \Rightarrow {{\omega }^{3}}=-\left( \omega +{{\omega }^{2}} \right) \\\ & \Rightarrow {{\omega }^{3}}=1 \\\ \end{aligned}
Therefore the value of the expression can be simplified by substituting the above values, so we get,
⇒E=1(22+2+1)+2(32+3+1)+....+(n−1)(n2+n+1)\Rightarrow E=1\left( {{2}^{2}}+2+1 \right)+2\left( {{3}^{2}}+3+1 \right)+....+\left( n-1 \right)\left( {{n}^{2}}+n+1 \right)
The above expression can be simplified as: -

& \Rightarrow E=\left( 2-1 \right)\left( {{2}^{2}}+\left( 2+1 \right) \right)+\left( 3-1 \right)\left( {{3}^{2}}+\left( 3+1 \right) \right)+....+\left( n-1 \right)\left( {{n}^{2}}+\left( n+1 \right) \right) \\\ & \Rightarrow E=\left( {{2}^{3}}-{{2}^{2}} \right)+\left( 2-1 \right)\left( 2+1 \right)+\left( {{3}^{3}}-{{3}^{2}} \right)+\left( 3-1 \right)\left( 3+1 \right)+....+\left( {{n}^{3}}-{{n}^{2}} \right)+\left( n-1 \right)\left( n+1 \right) \\\ \end{aligned}$$ Using the algebraic identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ we get, $$\Rightarrow E=\left( {{2}^{3}}-{{2}^{2}} \right)+\left( {{2}^{2}}-{{1}^{2}} \right)+\left( {{3}^{3}}-{{3}^{2}} \right)+\left( {{3}^{2}}-{{1}^{2}} \right)+....+\left( {{n}^{3}}-{{n}^{2}} \right)+\left( {{n}^{2}}-{{1}^{2}} \right)$$ Cancelling the like terms we get, $$\begin{aligned} & \Rightarrow E=\left( {{2}^{3}}-{{1}^{2}} \right)+\left( {{3}^{3}}-{{1}^{2}} \right)+....+\left( {{n}^{3}}-{{1}^{2}} \right) \\\ & \Rightarrow E=\left( {{2}^{3}}+{{3}^{3}}+.....+{{n}^{3}} \right)-\left( {{1}^{2}}+{{1}^{2}}+.....+{{1}^{2}} \right) \\\ & \Rightarrow E=\left( {{2}^{3}}+{{3}^{3}}+.....+{{n}^{3}} \right)-\left( 1+1+1+...+1 \right) \\\ \end{aligned}$$ Now, there are (n – 1) terms inside each bracket because the terms are starting from 2 and ending at n. So let us add ${{1}^{3}}$ and subtract 1 in the above expression which will have no effect on the value of the expression because ${{1}^{3}}=1$. So we get, $$\Rightarrow E=\left( {{1}^{3}}+{{2}^{3}}+{{3}^{3}}+.....+{{n}^{3}} \right)-\left( 1+1+1+1+...+1 \right)$$ Now, there are n terms in the above expression inside each bracket so we can write the expression in the summation form as: - $\begin{aligned} & \Rightarrow E=\sum\limits_{1}^{n}{\left( {{n}^{3}}-1 \right)} \\\ & \Rightarrow E=\sum\limits_{1}^{n}{\left( {{n}^{3}} \right)}-\sum\limits_{1}^{n}{\left( 1 \right)} \\\ \end{aligned}$ Using the formulas $\sum\limits_{1}^{n}{{{n}^{3}}}={{\left( \dfrac{n\left( n+1 \right)}{2} \right)}^{2}}$ and $\sum\limits_{1}^{n}{1}=n$ we get, $$\begin{aligned} & \Rightarrow E={{\left( \dfrac{n\left( n+1 \right)}{2} \right)}^{2}}-n \\\ & \therefore E=\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}-n \\\ \end{aligned}$$ **Hence, option (b) is the correct answer.** **Note:** You must remember all the formulas related to the cube roots of unity. Note that the value of $\omega $ is equal to $\dfrac{-1+\sqrt{3}i}{2}$ and that of $${{\omega }^{2}}$$ is equal to $\dfrac{-1-\sqrt{3}i}{2}$. Here $i$ id the imaginary number $\sqrt{-1}$. Remember the formulas of sum of first n natural numbers, sum of squares of first n natural numbers and the sum of cubes of first n natural numbers.