Questions & Answers

Question

Answers

A. $\dfrac{{ - n}}{{{{\left( {1 + \alpha } \right)}^2}}}$

B. $\dfrac{{ - n}}{{\left( {1 - \alpha } \right)}}$

C. \[\dfrac{{ - 2n}}{{\left( {1 - \alpha } \right)}}\]

D. \[\dfrac{{ - 2n}}{{{{\left( {1 - \alpha } \right)}^2}}}\]

Answer
Verified

Hint: Convert the given series into a finite GP & then find the sum suitably then use the properties of $\alpha$ which is the ${n^{th}}$ root of unity to get the final answer.

Complete step-by-step answer:

\[S = 1 + 2\alpha + 3{\alpha ^2} + ... + n{\alpha ^{n - 1}}\]â€¦..(1)

Multiplying both sides with $\alpha$, we get,

\[\alpha S = \alpha + 2{\alpha ^2} + 3{\alpha ^3} + ... + \left( {n - 1} \right){\alpha ^{n - 1}} + n{\alpha ^n}\] â€¦..(2)

Subtracting eqn (2) from (1)

\[S\left( {1 - \alpha } \right) = 1 + \alpha + {\alpha ^2} + {\alpha ^3} + ... + {\alpha ^{n - 1}} - n{\alpha ^n}\]

Using the formula of sum of G.P.,

Sum of G.P = \[{S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{\left( {1 - r} \right)}}\] , where n is the number of terms, r is the common ratio and a being the first term of G.P.

We get,

\[S\left( {1 - \alpha } \right) = \dfrac{{1 - {\alpha ^n}}}{{1 - \alpha }} - n{\alpha ^n}\]

Now \[{\alpha ^n} = 1\] since it is in the \[{n^{th}}\] root of unity.

Therefore,

\[S\left( {1 - \alpha } \right) = - n\]

\[S = \dfrac{{ - n}}{{\left( {1 - \alpha } \right)}}\]

Therefore, Option B is the correct answer.

Note: The given series sum is converted into a finite GP. The first term and the common ratio was found and thereby the sum of the GP was found. It was further simplified by using the property of $\alpha$.

Complete step-by-step answer:

\[S = 1 + 2\alpha + 3{\alpha ^2} + ... + n{\alpha ^{n - 1}}\]â€¦..(1)

Multiplying both sides with $\alpha$, we get,

\[\alpha S = \alpha + 2{\alpha ^2} + 3{\alpha ^3} + ... + \left( {n - 1} \right){\alpha ^{n - 1}} + n{\alpha ^n}\] â€¦..(2)

Subtracting eqn (2) from (1)

\[S\left( {1 - \alpha } \right) = 1 + \alpha + {\alpha ^2} + {\alpha ^3} + ... + {\alpha ^{n - 1}} - n{\alpha ^n}\]

Using the formula of sum of G.P.,

Sum of G.P = \[{S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{\left( {1 - r} \right)}}\] , where n is the number of terms, r is the common ratio and a being the first term of G.P.

We get,

\[S\left( {1 - \alpha } \right) = \dfrac{{1 - {\alpha ^n}}}{{1 - \alpha }} - n{\alpha ^n}\]

Now \[{\alpha ^n} = 1\] since it is in the \[{n^{th}}\] root of unity.

Therefore,

\[S\left( {1 - \alpha } \right) = - n\]

\[S = \dfrac{{ - n}}{{\left( {1 - \alpha } \right)}}\]

Therefore, Option B is the correct answer.

Note: The given series sum is converted into a finite GP. The first term and the common ratio was found and thereby the sum of the GP was found. It was further simplified by using the property of $\alpha$.

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