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Question: If \(\alpha \) is the \({n^{th}}\) root of unity, then \(1 + 2\alpha + 3{\alpha ^2} + ... + \,\,to\,...

If α\alpha is the nth{n^{th}} root of unity, then 1+2α+3α2+...+  to n1 + 2\alpha + 3{\alpha ^2} + ... + \,\,to\,n terms equal to
A. −n(1+α)2\dfrac{{ - n}}{{{{\left( {1 + \alpha } \right)}^2}}}
B. −n(1−α)\dfrac{{ - n}}{{\left( {1 - \alpha } \right)}}
C. −2n(1−α)\dfrac{{ - 2n}}{{\left( {1 - \alpha } \right)}}
D. −2n(1−α)2\dfrac{{ - 2n}}{{{{\left( {1 - \alpha } \right)}^2}}}

Explanation

Solution

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

Complete step-by-step answer:

S=1+2α+3α2+...+nαn−1S = 1 + 2\alpha + 3{\alpha ^2} + ... + n{\alpha ^{n - 1}}…..(1)

Multiplying both sides with α\alpha, we get,
αS=α+2α2+3α3+...+(n−1)αn−1+nαn\alpha S = \alpha + 2{\alpha ^2} + 3{\alpha ^3} + ... + \left( {n - 1} \right){\alpha ^{n - 1}} + n{\alpha ^n} …..(2)

Subtracting eqn (2) from (1)

S(1−α)=1+α+α2+α3+...+αn−1−nαnS\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 = Sn=a(1−rn)(1−r){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(1−α)=1−αn1−α−nαnS\left( {1 - \alpha } \right) = \dfrac{{1 - {\alpha ^n}}}{{1 - \alpha }} - n{\alpha ^n}

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

Therefore,
S(1−α)=−nS\left( {1 - \alpha } \right) = - n
S=−n(1−α)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 α\alpha.