Question

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

Answer 1

Hint: First analyze the repeating terms then use the property of the cube root of unity .After utilizing these formulae arrange in required format. Always remember the basic formulae of series for getting results rapidly .

Formula used: \[\ {
  1 + w + {w^2} = 0 \\
  w + {w^2} = - 1 \\
  {w^3} = 1 \\
  \sum\limits_{i = 1}^n {{n^3}} = {\left[ {\dfrac{{n(n + 1)}}{2}} \right]^2} = \dfrac{{{n^2}{{(n + 1)}^2}}}{4} \\
}\ \]

Complete step-by-step answer:
Given expression is \[1 \times (2 - w) \times (2 - {w^2}) + 2 \times (3 - w) \times (3 - {w^2}) + ............. + (n - 1) \times (n - w) \times (n - {w^2})\]
In this expression nth term can be written as \[{T_n}\]
\[{T_n} = (n - 1)(n - w)(n - {w^2})\]
The sum of all the term can be written as of following
\[S = \sum\limits_{i = 1}^n {(n - 1)(n - w)(n - {w^2})} \]
Now multiplying the terms we get
\[S = \]\[\sum\limits_{i = 1}^n {(n - 1)({n^2} - n{w^2}} - nw + {w^3})\]
\[S = \]\[\sum\limits_{i = 1}^n {(n - 1)({n^2} - n(w + {w^2}) + {w^3})} \] using the property of cube root of unity \[\ {
  w + {w^2} = - 1 \\
  {w^3} = 1 \\
\ } \]
\[S = \]\[\sum\limits_{i = 1}^n {(n - 1)({n^2} + n + 1)} \]
\[S = \]\[\sum\limits_{i = 1}^n {({n^3} + {n^2} + n} - {n^2} - n - 1)\]
further simplifying
\[S = \]\[\sum\limits_{i = 1}^n {({n^3} - 1)} \]=\[\sum\limits_{i = 1}^n {{n^3}} + \sum\limits_{i = 1}^n {( - 1)} \]=\[{\left[ {\dfrac{{n(n + 1)}}{2}} \right]^2} - n\]
\[S = \]\[\dfrac{{{n^2}({n^2} + 2n + 1) - 4n}}{4} = \dfrac{1}{4}n({n^3} + 2{n^2} + n - 4)\]=\[\dfrac{1}{4}n(n - 1)({n^2} + 3n + 4)\]
The solution obtained here is
\[S = \]\[\dfrac{1}{4}n(n - 1)({n^2} + 3n + 4)\]

Hence option (B) is the right answer.

Note: For solving the problem based on series remember the basic formulae of series
\[\sum\limits_{i = 1}^n {n = \dfrac{{n(n - 1)}}{2}} \] (sum of the first n natural numbers)
\[\sum\limits_{i = 1}^n {{n^2} = \dfrac{{n(n + 1)(2n + 1)}}{6}} \] (sum of the squares of the first n natural numbers)
\[\sum\limits_{i = 1}^n {{n^3}} = {\left[ {\dfrac{{n(n + 1)}}{2}} \right]^2} = \dfrac{{{n^2}{{(n + 1)}^2}}}{4} = {\left( {\sum\limits_{i = 1}^n n } \right)^2}\](sum of the cubes of first n natural numbers)
\[\sum\limits_{i = 1}^n {{n^4}} = \dfrac{n}{{30}}(n + 1)(2n + 1)(3{n^2} + 3n - 1)\] (sum of the fourth power of first n natural numbers)