# If $\alpha $is the nth root of unity, then $1 + 2\alpha + 3{\alpha ^2} + .....{\text{to n terms}}$ is equal to:

(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}}}$

Last updated date: 20th Mar 2023

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Answer

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Hint: Convert $1 + 2\alpha + 3{\alpha ^2} + .....{\text{to n terms}}$ in to a geometric series, apply the formula for sum of terms of G.P. and use ${\alpha ^n} = 1$

According to the question, $\alpha $ is the nth root of unity. So, we have:

$ \Rightarrow {\alpha ^n} = 1$

The expression given in the question is $1 + 2\alpha + 3{\alpha ^2} + .....{\text{to n terms}}$. Let it be $S$. Then we have:

\[

\Rightarrow S = 1 + 2\alpha + 3{\alpha ^2} + .....n{\alpha ^{n - 1}}{\text{,}} \\

\Rightarrow S\alpha = a + 2{\alpha ^2} + 3{\alpha ^3} + .....\left( {n - 1} \right){\alpha ^{n - 1}} + n{\alpha ^n}, \\

\Rightarrow S - S\alpha = 1 + a + {a^2} + .... + {a^{n - 1}} - n{a^n}, \\

\Rightarrow S\left( {1 - \alpha } \right) = 1 + a + {a^2} + .... + {a^{n - 1}} - n{a^n} \\

\]

We know that the sum of n terms of G.P. is $\dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$, using this, we’ll get:

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

We also know that ${\alpha ^n} = 1$, putting its value, we’ll get:

\[

\Rightarrow S\left( {1 - \alpha } \right) = \dfrac{{1 \times \left( {1 - 1} \right)}}{{\alpha - 1}} - n\left( 1 \right), \\

\Rightarrow S\left( {1 - \alpha } \right) = 0 - n, \\

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

\]

Substituting the value of $S$, we’ll get:

$ \Rightarrow 1 + 2\alpha + 3{\alpha ^2} + .....{\text{to n terms}} = - \dfrac{n}{{\left( {1 - \alpha } \right)}}$.

Thus, (B) is the correct option.

Note: If the G.P. consists of infinite terms then sum of its terms is:

$ \Rightarrow S = \dfrac{a}{{1 - r}}$, where r is the common ratio and a is the first term. But in such a case, for sum to be defined, the condition $0 < r < 1$ must hold.

According to the question, $\alpha $ is the nth root of unity. So, we have:

$ \Rightarrow {\alpha ^n} = 1$

The expression given in the question is $1 + 2\alpha + 3{\alpha ^2} + .....{\text{to n terms}}$. Let it be $S$. Then we have:

\[

\Rightarrow S = 1 + 2\alpha + 3{\alpha ^2} + .....n{\alpha ^{n - 1}}{\text{,}} \\

\Rightarrow S\alpha = a + 2{\alpha ^2} + 3{\alpha ^3} + .....\left( {n - 1} \right){\alpha ^{n - 1}} + n{\alpha ^n}, \\

\Rightarrow S - S\alpha = 1 + a + {a^2} + .... + {a^{n - 1}} - n{a^n}, \\

\Rightarrow S\left( {1 - \alpha } \right) = 1 + a + {a^2} + .... + {a^{n - 1}} - n{a^n} \\

\]

We know that the sum of n terms of G.P. is $\dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$, using this, we’ll get:

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

We also know that ${\alpha ^n} = 1$, putting its value, we’ll get:

\[

\Rightarrow S\left( {1 - \alpha } \right) = \dfrac{{1 \times \left( {1 - 1} \right)}}{{\alpha - 1}} - n\left( 1 \right), \\

\Rightarrow S\left( {1 - \alpha } \right) = 0 - n, \\

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

\]

Substituting the value of $S$, we’ll get:

$ \Rightarrow 1 + 2\alpha + 3{\alpha ^2} + .....{\text{to n terms}} = - \dfrac{n}{{\left( {1 - \alpha } \right)}}$.

Thus, (B) is the correct option.

Note: If the G.P. consists of infinite terms then sum of its terms is:

$ \Rightarrow S = \dfrac{a}{{1 - r}}$, where r is the common ratio and a is the first term. But in such a case, for sum to be defined, the condition $0 < r < 1$ must hold.

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