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Question: Let \( \alpha ,\beta \) denote the cube roots of unity other than 1 and \( \alpha \ne \beta \) . L...

Let α,β\alpha ,\beta denote the cube roots of unity other than 1 and α≠β\alpha \ne \beta .
Let S=∑n=0302(−1)n(αβ)nS = \sum\limits_{n = 0}^{302} {{{( - 1)}^n}{{\left( {\dfrac{\alpha }{\beta }} \right)}^n}} . Then the value of S is
A. Either −2ω- 2\omega or −2ω2- 2{\omega ^2}
B. Either −2ω- 2\omega or 2ω22{\omega ^2}
C. Either 2ω2\omega or −2ω2- 2{\omega ^2}
D. Either 2ω2\omega or 2ω22{\omega ^2}

Explanation

Solution

Start by considering a case where α=ω\alpha = \omega and β=ω2\beta = {\omega ^2} , find out the value of the expression given by putting different values of n from 0 to 302 . Use the properties of the cube root of unity to simplify the expression. Follow the same procedure for another case α=ω2\alpha = {\omega ^2} and β=ω\beta = \omega .

Complete step-by-step answer:
In order to find the value of S , We will have two cases
Case 1 :- Let α=ω\alpha = \omega and β=ω2\beta = {\omega ^2}
Now substituting the value of α\alpha and β\beta , we have
⇒S=∑n=0302(−1)n(ωω2)n ⇒S=∑n=0302(−1)n(ω−1)n  \Rightarrow S = \sum\limits_{n = 0}^{302} {{{( - 1)}^n}{{\left( {\dfrac{\omega }{{{\omega ^2}}}} \right)}^n}} \\\ \Rightarrow S = {\sum\limits_{n = 0}^{302} {\left( { - 1} \right)} ^n}{\left( {{\omega ^{ - 1}}} \right)^n} \\\
Now , Let us put the value of n from n = 0 to 302, we get
⇒1 - 1ω+1ω2−1ω3+.......................−1ω301+1ω302\Rightarrow {\text{1 - }}\dfrac{1}{\omega } + \dfrac{1}{{{\omega ^2}}} - \dfrac{1}{{{\omega ^3}}} + ....................... - \dfrac{1}{{{\omega ^{301}}}} + \dfrac{1}{{{\omega ^{302}}}}
Taking LCM and simplifying by taking 3 at a time in group, we have
⇒ω2−ω+1ω2−ω2−ω+1ω5.......................ω2−ω+1ω302\Rightarrow \dfrac{{{\omega ^2} - \omega + 1}}{{{\omega ^2}}} - \dfrac{{{\omega ^2} - \omega + 1}}{{{\omega ^5}}}.......................\dfrac{{{\omega ^2} - \omega + 1}}{{{\omega ^{302}}}}
Multiplying and divide by ω\omega , we get
⇒ω3−ω2+ωω3−ω3−ω2+ωω3.......................ω3−ω2+ωω303\Rightarrow \dfrac{{{\omega ^3} - {\omega ^2} + \omega }}{{{\omega ^3}}} - \dfrac{{{\omega ^3} - {\omega ^2} + \omega }}{{{\omega ^3}}}.......................\dfrac{{{\omega ^3} - {\omega ^2} + \omega }}{{{\omega ^{303}}}}
Since , it is cube root of unity ω3=1{\omega ^3} = 1 and 1+ω+ω2=01 + \omega + {\omega ^2} = 0
Hence , substituting the value in above equation , we have
⇒1−ω2+ω1−1−ω2+ω1.......................1−ω2+ω1\Rightarrow \dfrac{{1 - {\omega ^2} + \omega }}{1} - \dfrac{{1 - {\omega ^2} + \omega }}{1}.......................\dfrac{{1 - {\omega ^2} + \omega }}{1}
On further simplification , we have
⇒0 + .................. + 1 - ω2+ω (∵1+ω=−ω2)\Rightarrow {\text{0 + }}..................{\text{ + 1 - }}{\omega ^2} + \omega {\text{ }}\left( {\because 1 + \omega = - {\omega ^2}} \right)
∴−ω2−ω2=−2ω2\therefore - {\omega ^2} - {\omega ^2} = - 2{\omega ^2}

Case -2 :- Let α=ω2\alpha = {\omega ^2} and β=ω\beta = \omega
Now substituting the value of α\alpha and β\beta , we have
⇒S=∑n=0302(−1)n(ω2ω)n ⇒S=∑n=0302(−1)n(ω)n  \Rightarrow S = \sum\limits_{n = 0}^{302} {{{( - 1)}^n}{{\left( {\dfrac{{{\omega ^2}}}{\omega }} \right)}^n}} \\\ \Rightarrow S = {\sum\limits_{n = 0}^{302} {\left( { - 1} \right)} ^n}{\left( \omega \right)^n} \\\
Now , Let us put the value of n from n = 0 to 302, we get
⇒1 - ω+ω2−ω3+.......................−ω301+ω302\Rightarrow {\text{1 - }}\omega + {\omega ^2} - {\omega ^3} + ....................... - {\omega ^{301}} + {\omega ^{302}}
Since , it is cube root of unity ω3=1{\omega ^3} = 1 and 1+ω+ω2=01 + \omega + {\omega ^2} = 0
Hence , substituting the value in above equation , we have
⇒1 - ω+ω2−1+ω−ω2+.......................1 - ω+ω2\Rightarrow {\text{1 - }}\omega + {\omega ^2} - 1 + \omega - {\omega ^2} + .......................1{\text{ - }}\omega + {\omega ^2}
On further simplification , we have
⇒0 + .................. + 1 - ω+ω2 (∵1+ω2=−ω)\Rightarrow {\text{0 + }}..................{\text{ + 1 - }}\omega + {\omega ^2}{\text{ }}\left( {\because 1 + {\omega ^2} = - \omega } \right)
∴−ω−ω=−2ω\therefore - \omega - \omega = - 2\omega

So, the correct answer is “Option A”.

Note: Similar question can be solved by using the properties of the cube root of unity. Students must also know the value of all the different powers of iota (i) , Attention must be given while substituting the values and simplifying as it might give wrong answers if mistaken or missed.