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

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Last updated date: 24th Jul 2024
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Answer
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Hint:Assume the given expression as E. Multiply the terms present in the expression to form a general pattern. Use the formulas $1+\omega +{{\omega }^{2}}=0$ and ${{\omega }^{3}}=1$ to simplify the expression. Now, form the summation series by cancelling the common terms and use the formulas $\sum\limits_{1}^{n}{{{n}^{3}}}={{\left( \dfrac{n\left( n+1 \right)}{2} \right)}^{2}}$ and $\sum\limits_{1}^{n}{1}=n$ to get the answer. Use the algebraic identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ for the simplification.

Complete step-by-step solution:
Here we have been provided with the 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 imaginary cube root of unity and we are asked to find its value. Let us assume the expression as E, so we have,
$\Rightarrow E=1\left( 2-\omega \right)\left( 2-{{\omega }^{2}} \right)+2\left( 3-\omega \right)\left( 3-{{\omega }^{2}} \right)+....+\left( n-1 \right)\left( n-\omega \right)\left( n-{{\omega }^{2}} \right)$
Now, multiplying the expression in each term we get,
$\Rightarrow E=1\left( {{2}^{2}}-2\left( \omega +{{\omega }^{2}} \right)+{{\omega }^{3}} \right)+2\left( {{3}^{2}}-3\left( \omega +{{\omega }^{2}} \right)+{{\omega }^{3}} \right)+....+\left( n-1 \right)\left( {{n}^{2}}-n\left( \omega +{{\omega }^{2}} \right)+{{\omega }^{3}} \right)$
We know that $1+\omega +{{\omega }^{2}}=0$, so we get $\omega +{{\omega }^{2}}=-1$. Also, multiplying both the sides of the expression $1+\omega +{{\omega }^{2}}=0$ with $\omega $ we get,
$\begin{align}
  & \Rightarrow \omega +{{\omega }^{2}}+{{\omega }^{3}}=0 \\
 & \Rightarrow {{\omega }^{3}}=-\left( \omega +{{\omega }^{2}} \right) \\
 & \Rightarrow {{\omega }^{3}}=1 \\
\end{align}$
Therefore the value of the expression can be simplified by substituting the above values, so we get,
\[\Rightarrow E=1\left( {{2}^{2}}+2+1 \right)+2\left( {{3}^{2}}+3+1 \right)+....+\left( n-1 \right)\left( {{n}^{2}}+n+1 \right)\]
The above expression can be simplified as: -
\[\begin{align}
  & \Rightarrow E=\left( 2-1 \right)\left( {{2}^{2}}+\left( 2+1 \right) \right)+\left( 3-1 \right)\left( {{3}^{2}}+\left( 3+1 \right) \right)+....+\left( n-1 \right)\left( {{n}^{2}}+\left( n+1 \right) \right) \\
 & \Rightarrow E=\left( {{2}^{3}}-{{2}^{2}} \right)+\left( 2-1 \right)\left( 2+1 \right)+\left( {{3}^{3}}-{{3}^{2}} \right)+\left( 3-1 \right)\left( 3+1 \right)+....+\left( {{n}^{3}}-{{n}^{2}} \right)+\left( n-1 \right)\left( n+1 \right) \\
\end{align}\]
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,
\[\begin{align}
  & \Rightarrow E=\left( {{2}^{3}}-{{1}^{2}} \right)+\left( {{3}^{3}}-{{1}^{2}} \right)+....+\left( {{n}^{3}}-{{1}^{2}} \right) \\
 & \Rightarrow E=\left( {{2}^{3}}+{{3}^{3}}+.....+{{n}^{3}} \right)-\left( {{1}^{2}}+{{1}^{2}}+.....+{{1}^{2}} \right) \\
 & \Rightarrow E=\left( {{2}^{3}}+{{3}^{3}}+.....+{{n}^{3}} \right)-\left( 1+1+1+...+1 \right) \\
\end{align}\]
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: -
$\begin{align}
  & \Rightarrow E=\sum\limits_{1}^{n}{\left( {{n}^{3}}-1 \right)} \\
 & \Rightarrow E=\sum\limits_{1}^{n}{\left( {{n}^{3}} \right)}-\sum\limits_{1}^{n}{\left( 1 \right)} \\
\end{align}$
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,
\[\begin{align}
  & \Rightarrow E={{\left( \dfrac{n\left( n+1 \right)}{2} \right)}^{2}}-n \\
 & \therefore E=\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}-n \\
\end{align}\]
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.