Question

Let alpha, and beta denote the cube root of unity other than 1 and alpha not equal beta. If \[S = {\sum\limits_{n = 0}^{302} {{{\left( { - 1} \right)}^n}\left( {\dfrac{\alpha }{\beta }} \right)} ^n}...\left( 1 \right)\]. Then the value of S is

Answer 1

Hint: In this question, we need to find the value of S. For that, we know that \[1,\omega ,{\omega ^2}\] are the three cube roots of unity then assume that \[\beta = \omega \] and \[\alpha = {\omega ^2}\] then substitute this value in S which is given and simplifying it to get the desired result.

Formula used:
We have used the following formulas:
 \[
  {S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}} \\
  Where\,\, \\
  a = first\,term \\
  r = common\,ratio \\
  n = number\,of\,terms
 \]

Complete step-by-step solution:
Given that \[S = {\sum\limits_{n = 0}^{302} {{{\left( { - 1} \right)}^n}\left( {\dfrac{\alpha }{\beta }} \right)} ^n}...\left( 1 \right)\]
Now we know that there are three cube roots of unity \[1,\omega ,{\omega ^2}\]
Now let us assume that \[\beta = \omega \] and \[\alpha = {\omega ^2}\]
Now substitute these values in equation (1):
\[
  S = {\sum\limits_{n = 0}^{302} {{{\left( { - 1} \right)}^n}\left( {\dfrac{{{\omega ^2}}}{\omega }} \right)} ^n} \\
  S = {\sum\limits_{n = 0}^{302} {{{\left( { - 1} \right)}^n}\left( \omega \right)} ^n}
 \]
Now When \[n = 0\]
\[
  S = {\left( { - 1} \right)^0}{\left( \omega \right)^0} \\
   = 1
 \]
When \[n = 1\]
\[
  S = {\left( { - 1} \right)^1}{\left( \omega \right)^1} \\
   = \left( { - 1} \right)\left( \omega \right) \\
   = - \omega
 \]
When \[n = 2\]
\[
  S = {\left( { - 1} \right)^2}{\left( \omega \right)^2} \\
   = \left( 1 \right)\left( {{\omega ^2}} \right) \\
   = {\omega ^2}
 \]
When \[n = 3\]
\[
  S = {\left( { - 1} \right)^3}{\left( \omega \right)^3} \\
   = \left( { - 1} \right)\left( {{\omega ^3}} \right) \\
   = - {\omega ^3}
 \]
When \[n = 4\]
\[
  S = {\left( { - 1} \right)^4}{\left( \omega \right)^4} \\
   = \left( 1 \right)\left( {{\omega ^4}} \right) \\
   = {\omega ^4}
 \]
When \[n = 302\]
\[
  S = {\left( { - 1} \right)^{302}}{\left( \omega \right)^{302}} \\
   = \left( 1 \right)\left( {{\omega ^{302}}} \right) \\
   = {\omega ^{302}}
 \]
Therefore, the series becomes
\[S = 1 - \omega + {\omega ^2} - {\omega ^3} + {\omega ^4} + ....... + {\omega ^{302}}\]
Now we can see that the above series is a G.P series because it has a constant common factor as it increases. Therefore,
 \[
  a = 1 \\
  r = \dfrac{b}{a} \\
   = \dfrac{{ - \omega }}{1} \\
   = \omega
 \]
We know that \[{S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}\]
Now substitute all the values in the formula:
\[
  {S_n} = \dfrac{{1\left( {1 - {{\left( { - \omega } \right)}^{303}}} \right)}}{{1 + \omega }} \\
   = \dfrac{{1 + {\omega ^{303}}}}{{1 + \omega }} \\
 \]
We know the property that \[{\omega ^{3n}} = 1,\,{\omega ^{3n + 1}} = \omega ,\,\,{\omega ^{3n + 2}} = {\omega ^2}\]
By substituting the value from this property, we get
\[
  {S_n} = \dfrac{{1 + 1}}{{1 + \omega }} \\
   = \dfrac{2}{{1 + \omega }}
 \]
To simplifying the above equation, we add and subtract \[{\omega ^2}\] in the denominator:
\[{S_n} = \dfrac{2}{{1 + \omega - {\omega ^2} + {\omega ^2}}}\]
We know that \[1 + \omega + {\omega ^2} = 0\]
By substituting this value in above, we get
\[
  {S_n} = - \dfrac{2}{{{\omega ^2}}} \\
   = - \dfrac{2}{{{\omega ^2}}} \times \dfrac{\omega }{\omega }\left( {rationalizing} \right) \\
   = \dfrac{{ - 2\omega }}{{{\omega ^3}}} \\
   = - 2\omega \left[ {\because {\omega ^3} = 1} \right]
 \]
Hence, the value of S is \[ - 2\omega \]

Note: Students must be very careful when solving the summation of the series because a calculation error may occur, and they must be familiar with the sum of the series formula which is
\[
  {S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}} \\
  Where\,\, \\
  a = first\,term \\
  r = common\,ratio \\
  n = number\,of\,terms
 \]