Question

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

Answer 1

Simplify the relation of the sequence and use the identities of cube root of unity . Then form the general term representing the sequence and use the general term in order to find out the sum of the required sequence.

Complete step-by-step answer :
Given
$1 \cdot (2 - \omega )(2 - {\omega ^2}) + {\text{ }}2 \cdot (3 - \omega )(3 - {\omega ^2}) + \ldots \ldots \ldots \ldots .. + {\text{ }}\left( {n - 1} \right)(n - \omega )(n - {\omega ^2})$
Let us start by opening the brackets and obtain a homogeneous equation. We get
$1 \cdot (4 - 2\omega - 2{\omega ^2} + {\omega ^3}) + {\text{ }}2 \cdot (9 - 3\omega - 3{\omega ^2} + {\omega ^3}) + \ldots \ldots \ldots \ldots .. + {\text{till }}\left( {n - 1} \right)terms \\\ 1 \cdot [4 - 2(\omega + {\omega ^2}) + {\omega ^3}] + {\text{ }}2 \cdot [9 - 3(\omega + {\omega ^2}) + {\omega ^3}] + \ldots \ldots \ldots \ldots .. + {\text{till }}\left( {n - 1} \right)terms \\\ \\\$
Now , We know that
$\omega + {\omega ^2} = - 1{\text{ and }}{\omega ^3} = 1.{\text{ Substituting these values ,we get}} \\\ 1 \cdot [4 - 2( - 1) + 1] + {\text{ }}2 \cdot [9 - 3( - 1) + 1] + \ldots \ldots \ldots \ldots .. + {\text{till }}\left( {n - 1} \right)terms \\\ \Rightarrow 1 \cdot [4 + 2 + 1] + {\text{ }}2 \cdot [9 + 3 + 1] + \ldots \ldots \ldots \ldots .. + {\text{till }}\left( {n - 1} \right)terms \\\ \Rightarrow 1 \cdot [7] + 2 \cdot [13] + ............ + {\text{ till }}\left( {n - 1} \right)terms \\\$
Let us try to find out the general term of the sequence.
$1 \cdot [7] + 2 \cdot [13] + 3 \cdot [21]............ + {\text{ till }}\left( {n - 1} \right)terms$
By observing the sequence we find that first part of the sequence is 1,2,3,4 …… till (n-1) terms, It’s general term is ‘x’ and the second part of the sequence in 7,13,21,….. till (n-1) terms , So we will try to find it’s general term.
$S = 7 + 13 + 21 + ................... + {r_x} \to eqn(a) \\\ S = 0 + 7 + 13 + 21 + .............. + {r_{x - 1}} + {r_x} \to eqn(b) \\\$
Subtracting equation b from equation a , by following the rule of subtracting first term from second and so on….
$0 = 7 + 6 + 8 + ................ + ({r_x} - {r_{x - 1}}) - {r_x} \\\ {r_x} = 7 + [6 + 8 + 10........ + ({r_x} - {r_{x - 1}})] \\\$
If we observe here 6,8,10 form an A.P. with common difference(d) = 2 and number of terms = (x-1). Applying the sum of first n natural number i.e.$= \dfrac{n}{2}[2a + (n - 1)d]$
{r_x} = 7 + \\{ \dfrac{{(x - 1)}}{2}[2 \times 6 + (x - 2) \times 2]\\} \\\ \Rightarrow {r_x} = 7 + \left\\{ {\dfrac{{(x - 1)}}{2}[12 + 2x - 4]} \right\\} \\\ \Rightarrow {r_x} = 7 + \left\\{ {\dfrac{{(x - 1)}}{2}[8 + 2x]} \right\\} \\\ \Rightarrow {r_x} = 7 + (x - 1)(4 + x) \\\ \Rightarrow {r_x} = 7 + 4x + {x^2} - 4 - x \\\ \Rightarrow {r_x} = 3 + 3x + {x^2} \\\
By splitting the middle term we get
${r_x} = {x^2} + 2x + x + 2 + 1 \\\ {r_x} = (x + 2)(x + 1) + 1 \\\$
So the final general term formed by the first and second sequence together will be as follows
${{\text{T}}_x} = x[(x + 1)(x + 2) + 1]{\text{ for }}x{\text{ = 1 to}}(n - 1)$
This can also be written as
$\Rightarrow {\text{ }}{{\text{T}}_x} = {x^3}{\text{ + 3}}{{\text{x}}^2}{\text{ + 3x for }}x{\text{ = 1 to}}(n - 1)$
Now ,we know the sum of sequence can be derived from general term as
${S_x} = \sum\limits_1^{n - 1} {{T_x}} \\\ \therefore S = \sum {{x^3}} + 3 \cdot \sum {{x^2}} + 3 \cdot \sum x {\text{ for }}x = 1{\text{ to (n - 1)}} \\\$
We know
$\sum {{x^3}} = {\text{ [}}\dfrac{{n(n + 1)}}{2}{]^2},\sum {{x^2}} = \dfrac{{n(n + 1)(2n + 1)}}{6},\sum x = \dfrac{{n(n + 1)}}{2} \\\ {\text{Here }}n{\text{ }} = {\text{ }}n{\text{ }} - 1 \\\ \Rightarrow {[\dfrac{{n(n - 1)}}{2}]^2} + 3[\dfrac{{n(n - 1)(2n - 1)}}{6}] + 3[\dfrac{{n(n - 1)}}{2}] \\\ \\\$
On Simplifying this , we get
$\Rightarrow \dfrac{1}{4}n(n - 1)({n^2} + 3n + 4)$
Therefore , Option B is the correct option.

Note : All the identities of the cube root of unity and formulas for the sum of first natural numbers , their squares and cubes must be known for solving such similar questions. Attention must be given while substituting the values and also while forming the general term.