Question

# If $\alpha$ is the ${n^{th}}$ root of unity, then $1 + 2\alpha + 3{\alpha ^2} + ... + \,\,to\,n$ terms equal toA. $\dfrac{{ - n}}{{{{\left( {1 + \alpha } \right)}^2}}}$B. $\dfrac{{ - n}}{{\left( {1 - \alpha } \right)}}$C. $\dfrac{{ - 2n}}{{\left( {1 - \alpha } \right)}}$D. $\dfrac{{ - 2n}}{{{{\left( {1 - \alpha } \right)}^2}}}$

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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.

$S = 1 + 2\alpha + 3{\alpha ^2} + ... + n{\alpha ^{n - 1}}$…..(1)

Multiplying both sides with $\alpha$, we get,
$\alpha S = \alpha + 2{\alpha ^2} + 3{\alpha ^3} + ... + \left( {n - 1} \right){\alpha ^{n - 1}} + n{\alpha ^n}$ …..(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 $\alpha$.