Question

The value of the expression $1.\left( 2-\omega \right) \left( 2-\omega^{2} \right) +2.\left( 3-\omega \right) \left( 3-\omega^{2} \right) +\cdots +\left( n-1\right) \left( n-\omega \right) \left( n-\omega^{2} \right)$, where $\omega$ is an imaginary cube root of unity, is
A) $\dfrac{1}{2} \left( n-1\right) n\left( n^{2}+3n+4\right)$
B) $\dfrac{1}{4} \left( n-1\right) n\left( n^{2}+3n+4\right)$
C) $\dfrac{1}{2} \left( n+1\right) n\left( n^{2}+3n+4\right)$
D) $\dfrac{1}{4} \left( n+1\right) n\left( n^{2}+3n+4\right)$

Answer 1

Hint: In this question it is given that we have to find the value of the expansion $1.\left( 2-\omega \right) \left( 2-\omega^{2} \right) +2.\left( 3-\omega \right) \left( 3-\omega^{2} \right) +\cdots +\left( n-1\right) \left( n-\omega \right) \left( n-\omega^{2} \right)$. So to find the solution we have to use the properties of $\omega$, as we know that $\omega$ is the imaginary cube root of 1, then we can write,
$\omega^{3} =1$................(1)
& $\omega +\omega^{2} =-1$………(2)

Complete step-by-step solution:
The given expression can be written as,
S=$1.\left( 2-\omega \right) \left( 2-\omega^{2} \right) +2.\left( 3-\omega \right) \left( 3-\omega^{2} \right) +\cdots +\left( n-1\right) \left( n-\omega \right) \left( n-\omega^{2} \right)$
=$\left( 2-1\right) \left( 2-\omega \right) \left( 2-\omega^{2} \right) +\left( 3-1\right) \left( 3-\omega \right) \left( 3-\omega^{2} \right) +\cdots +\left( n-1\right) \left( n-\omega \right) \left( n-\omega^{2} \right)$
=$\displaystyle \sum^{n}_{k=2} \left( k-1\right) \left( k-\omega \right) \left( k-\omega^{2} \right)$......(3)[ applying the summation from k=2 to n]
=$\displaystyle \sum^{n}_{k=2} \left( k-1\right) \left( k^{2}-k\omega -k\omega^{2} +\omega^{3} \right)$
=\displaystyle \sum^{n}_{k=2} \left( k-1\right) \left\\{ k^{2}-k\left( \omega +\omega^{2} \right) +\omega^{3} \right\\}
=\displaystyle \sum^{n}_{k=2} \left( k-1\right) \left\\{ k^{2}-k\left( -1\right) +1\right\\} [ by using (1) and (2)]
=\displaystyle \sum^{n}_{k=2} \left( k-1\right) \left\\{ k^{2}+k+1\right\\}
=$\displaystyle \sum^{n}_{k=2} \left( k^{3}-1\right)$ [ $\because \left( a-b\right) \left( a^{2}+ab+b^{2}\right) =a^{3}-b^{3}$]
Now by expanding the summation we get,
S=$\left( 2^{3}-1\right) +\left( 3^{3}-1\right) +\cdots +\left( n^{3}-1\right)$
=$\left( 2^{3}+3^{3}+\cdots +n^{3}\right) +\left( n-1\right)$ [ since, in the $2^{nd}$ part (n-1) times 1 are there]
=$\left( 1+2^{3}+3^{3}+\cdots +n^{3}\right) +n$
As we know that, the summation of the cube of first ‘n’ positive integer is \left\\{ \dfrac{n\left( n+1\right) }{2} \right\\}^{2} .
So by this we can write,
S=\left\\{ \dfrac{n\left( n+1\right) }{2} \right\\}^{2} +n
=$\dfrac{n^{2}\left( n^{2}+2n+1\right) }{4} +n$
=\dfrac{n}{4} \left\\{ n\left( n^{2}+2n+1\right) +4\right\\} [taking $\dfrac{n}{4}$ common]
=$\dfrac{n}{4} \left( n^{3}+2n^{2}+n+4\right)$
Now solving the cubic equation by factorising, we get,
S=$\dfrac{n}{4} \left( n^{3}-n^{2}+3n^{2}-3n+4n+4\right)$
Now we are going to take common $n^{2}$ from $1^{st}$ and $2^{nd}$ term, 3n from $3^{rd}$ and $4^{th}$ term, 4 from $5^{th}$ and $6^{th}$ term.
Therefore, we get,
S=\dfrac{n}{4} \left\\{ n^{2}\left( n-1\right) +3n\left( n-1\right) +4\left( n-1\right) \right\\}
=$\dfrac{n}{4} \left( n-1\right) \left( n^{2}+3n+4\right)$ [ by taking (n - 1) common]
=$\dfrac{1}{4} \left( n-1\right) n\left( n^{2}+3n+4\right)$
Hence the correct option is option B.

Note: While solving any series try to write it using the summation which we have used in step (3), this will shorten the calculation process. Also while writing any series as summation you have to keep in mind that each and every term of the series following a certain structure or else what you can do , you can find its $n^{th}$ term i.e, $t_{n}=\left( n-1\right) \left( n-\omega \right) \left( n-\omega^{2} \right)$ and then use summation notation replacing n by k and writing it as $\displaystyle \sum^{n}_{k=2} \left( k-1\right) \left( k-\omega \right) \left( k-\omega^{2} \right)$.