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Question: If \[\alpha \] not equal 1 is any \(n^{th}\) root of unity then \[S = 1 + 3\alpha + 5{\alpha ^2} + ....

If α\alpha not equal 1 is any nthn^{th} root of unity then S=1+3α+5α2+...nS = 1 + 3\alpha + 5{\alpha ^2} + ...n terms equals:
A. 2n1−α\dfrac{{2n}}{{1 - \alpha }}
B. −2n1−α\dfrac{{ - 2n}}{{1 - \alpha }}
C. n1−α\dfrac{n}{{1 - \alpha }}
D. −n1−α\dfrac{{ - n}}{{1 - \alpha }}

Explanation

Solution

Here in the given question we are given series in which we need to find the summation of the series, to solve this we need to get the formulae for the summation of series, and this we can get from writing the general term for the given terms, in order to get the related formulae, here the general term would be:
⇒Tr=(2r−1)ar−1\Rightarrow {T_r} = (2r - 1){a^{r - 1}}
Formulae of summation of the above series:
⇒S=∑r=1n(2r−1)αr−1=2∑r=1nrαr−1−∑r=1nαr−1\Rightarrow S = \sum\limits_{r = 1}^n {{{\left( {2r - 1} \right)}^{{\alpha ^{r - 1}}}}} = 2\sum\limits_{r = 1}^n r {\alpha ^{r - 1}} - \sum\limits_{r = 1}^n {{\alpha ^{r - 1}}}
Formulae Used: Summation for series with the general term as:
⇒Tr=(2r−1)ar−1\Rightarrow {T_r} = (2r - 1){a^{r - 1}}
⇒S=∑r=1n(2r−1)αr−1=2∑r=1nrαr−1−∑r=1nαr−1\Rightarrow S = \sum\limits_{r = 1}^n {{{\left( {2r - 1} \right)}^{{\alpha ^{r - 1}}}}} = 2\sum\limits_{r = 1}^n r {\alpha ^{r - 1}} - \sum\limits_{r = 1}^n {{\alpha ^{r - 1}}}

Complete step by step answer:
Here in the given question we need to solve with the help of the summation of series formula, here we need to put the values according to the term given in the question and then simply solve further in order to get the solution, on solving we get-
We have-

⇒S1=∑r=1nrαr−1 ⇒S2=∑r=1nαr−1  \Rightarrow {S_1} = \sum\limits_{r = 1}^n r {\alpha ^{r - 1}} \\\ \Rightarrow {S_2} = \sum\limits_{r = 1}^n {{\alpha ^{r - 1}}} \\\

Now we can write-
⇒S=2∑r=1nrαr−1−∑r=1nαr−1=2S1−S2\Rightarrow S = 2\sum\limits_{r = 1}^n r {\alpha ^{r - 1}} - \sum\limits_{r = 1}^n {{\alpha ^{r - 1}}} = 2{S_1} - {S_2}
And
⇒S2=αS1\Rightarrow {S_2} = \alpha {S_1}

⇒S1=1+2α+3α2+4α3+...+nαn−1 ⇒S2=α+2α2+3α3+4α4+...+nαn=αS1 ⇒S1−S2=(1+2α+3α2+4α3+...+nαn−1)−(α+2α2+3α3+4α4+...+nαn) ⇒S1−αS1=(1+α+α2+...+αn−1)−nαn ⇒S1(1−α)=(1+α+α2+...+αn−1)−nαn  \Rightarrow {S_1} = 1 + 2\alpha + 3{\alpha ^2} + 4{\alpha ^3} + ... + n{\alpha ^{n - 1}} \\\ \Rightarrow {S_2} = \alpha + 2{\alpha ^2} + 3{\alpha ^3} + 4{\alpha ^4} + ... + n{\alpha ^n} = \alpha {S_1} \\\ \Rightarrow {S_1} - {S_2} = \left( {1 + 2\alpha + 3{\alpha ^2} + 4{\alpha ^3} + ... + n{\alpha ^{n - 1}}} \right) - \left( {\alpha + 2{\alpha ^2} + 3{\alpha ^3} + 4{\alpha ^4} + ... + n{\alpha ^n}} \right) \\\ \Rightarrow {S_1} - \alpha {S_1} = \left( {1 + \alpha + {\alpha ^2} + ... + {\alpha ^{n - 1}}} \right) - n{\alpha ^n} \\\ \Rightarrow {S_1}\left( {1 - \alpha } \right) = \left( {1 + \alpha + {\alpha ^2} + ... + {\alpha ^{n - 1}}} \right) - n{\alpha ^n} \\\

Here we got a series of GP in above equation, using summation of series of GP we get:
⇒1+α+α2+...+αn−1=(1−αn)(1−α)\Rightarrow 1 + \alpha + {\alpha ^2} + ... + {\alpha ^{n - 1}} = \dfrac{{\left( {1 - {\alpha ^n}} \right)}}{{\left( {1 - \alpha } \right)}}
Putting value we get:

⇒S1(1−α)=(1−αn)(1−α)−nαn ⇒S1=(1−αn)(1−α)×(1−α)−nαn(1−α) ⇒S1=(1−αn)(1−α)2−nαn(1−α)  \Rightarrow {S_1}\left( {1 - \alpha } \right) = \dfrac{{\left( {1 - {\alpha ^n}} \right)}}{{\left( {1 - \alpha } \right)}} - n{\alpha ^n} \\\ \Rightarrow {S_1} = \dfrac{{\left( {1 - {\alpha ^n}} \right)}}{{\left( {1 - \alpha } \right) \times \left( {1 - \alpha } \right)}} - \dfrac{{n{\alpha ^n}}}{{\left( {1 - \alpha } \right)}} \\\ \Rightarrow {S_1} = \dfrac{{\left( {1 - {\alpha ^n}} \right)}}{{{{\left( {1 - \alpha } \right)}^2}}} - \dfrac{{n{\alpha ^n}}}{{\left( {1 - \alpha } \right)}} \\\

Here we know α\alpha , is the nthn^{th} root of unity, hence αn=1{\alpha ^n} = 1 and,
⇒1+α+α2+...+αn−1=0\Rightarrow 1 + \alpha + {\alpha ^2} + ... + {\alpha ^{n - 1}} = 0
⇒S1=(1−1)(1−α)2−n(1)(1−α)=0−n(1−α)=−n(1−α)\Rightarrow {S_1} = \dfrac{{\left( {1 - 1} \right)}}{{{{\left( {1 - \alpha } \right)}^2}}} - \dfrac{{n\left( 1 \right)}}{{\left( {1 - \alpha } \right)}} = 0 - \dfrac{n}{{\left( {1 - \alpha } \right)}} = \dfrac{{ - n}}{{\left( {1 - \alpha } \right)}}
Hence
⇒S2=αS1=∑r=1nαr−1=1+α+α2+...+αn−1=0\Rightarrow {S_2} = \alpha {S_1} = \sum\limits_{r = 1}^n {{\alpha ^{r - 1}}} = 1 + \alpha + {\alpha ^2} + ... + {\alpha ^{n - 1}} = 0
Now,
⇒S=2S1−S2=2S1−0=2S1=2(−n(1−α))\Rightarrow S = 2{S_1} - {S_2} = 2{S_1} - 0 = 2{S_1} = 2\left( {\dfrac{{ - n}}{{\left( {1 - \alpha } \right)}}} \right)
Here we get the solution for the summation of the given series.

So, the correct answer is “Option A”.

Note: Here in the given series, the given series is incomplete hence we need to first calculate for the whole series, then with the given condition we have found the related value, and after adjustment we reach the final solution for the summation of the given series.