Courses
Courses for Kids
Free study material
Free LIVE classes
More

# 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
Total views: 306.9k
Views today: 6.85k
Verified
306.9k+ views
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.