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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)\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 is:
A. [n(n+1)2]2{\left[ {\dfrac{{n\left( {n + 1} \right)}}{2}} \right]^2}
B. [n(n+1)2]2−n{\left[ {\dfrac{{n\left( {n + 1} \right)}}{2}} \right]^2} - n
C. [n(n+1)2]2+n{\left[ {\dfrac{{n\left( {n + 1} \right)}}{2}} \right]^2} + n
D. None of these

Explanation

Solution

Hint: First of all, write the mth{m^{th}} term of the given series. And then, use the property of ω\omega to simplify the given condition. Next, add and subtract mm to the expression and use the property 1+ω+ω2=01 + \omega + {\omega ^2} = 0 to simplify the mth{m^{th}} term of given expression. Hence, use the direct formula of summation to find the required answer.

Complete step by step answer:

The mth{m^{th}} term of the given series 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) can be written as (m−1)(m−ω)(m−ω2)\left( {m - 1} \right)\left( {m - \omega } \right)\left( {m - {\omega ^2}} \right).
We can simplify the expression for the mth{m^{th}} term of the series.
(m−1)(m2−mω2−mω+ω3)\left( {m - 1} \right)\left( {{m^2} - m{\omega ^2} - m\omega + {\omega ^3}} \right)
It is known that ω3=1{\omega ^3} = 1 since ω\omega is the cube root of the unity.
Therefore the expression (m−1)(m2−mω2−mω+ω3)\left( {m - 1} \right)\left( {{m^2} - m{\omega ^2} - m\omega + {\omega ^3}} \right) becomes
(m−1)(m2−mω2−mω+1)\left( {m - 1} \right)\left( {{m^2} - m{\omega ^2} - m\omega + 1} \right)
Adding mmand subtracting mm in the second factor of the expression and simplifying, we get
=(m−1)(m2−mω2−mω−m+m+1) =(m−1)(m2−(ω2+ω+1)m+m+1)  = \left( {m - 1} \right)\left( {{m^2} - m{\omega ^2} - m\omega - m + m + 1} \right) \\\ = \left( {m - 1} \right)\left( {{m^2} - \left( {{\omega ^2} + \omega + 1} \right)m + m + 1} \right) \\\
It is known, the property of ω\omega , that is 1+ω+ω2=01 + \omega + {\omega ^2} = 0. Thus the expression (m−1)(m2−(ω2+ω+1)m+m+1)\left( {m - 1} \right)\left( {{m^2} - \left( {{\omega ^2} + \omega + 1} \right)m + m + 1} \right) reduces to
=(m−1)(m2+m+1)= \left( {m - 1} \right)\left( {{m^2} + m + 1} \right)
Simplifying the expression for the mth{m^{th}} term of the series, we get
=(m−1)(m2+m+1) =m3+m2+m−m2−m−1 =m3−1  = \left( {m - 1} \right)\left( {{m^2} + m + 1} \right) \\\ = {m^3} + {m^2} + m - {m^2} - m - 1 \\\ = {m^3} - 1 \\\
The sum of the given series 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) can be evaluated by taking the summation of the mth{m^{th}} term of the series from m=1m = 1 to m=nm = n.
Thus the sum of given series is
∑m=1nm3−1\sum\limits_{m = 1}^n {{m^3} - 1}
Solving the expression ∑m=1nm3−1\sum\limits_{m = 1}^n {{m^3} - 1} , we get
=∑m=1nm3−∑m=1n1 =(n(n+1)2)2−n  = \sum\limits_{m = 1}^n {{m^3}} - \sum\limits_{m = 1}^n 1 \\\ = {\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2} - n \\\
The sum of the given series 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 (n(n+1)2)2−n{\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2} - n.
Thus the option B is the correct answer.

Note: The summation of x3{x^3} over the range 1 to nn is (n(n+1)2)2{\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2}. The summation of 1 for nntimes is equal to nn. Similarly, The summation of x2{x^2} over the range 1 to nn is n(n+1)(2n+1)6\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}. And, summation of xx over the range 1 to nn is n(n+1)2\dfrac{{n\left( {n + 1} \right)}}{2}.