Question

If $\alpha$ is a root of the equation $x^2-x+1=0$ then $(\alpha + \frac{1}{\alpha}) + (\alpha^2 + \frac{1}{\alpha^2}) + (\alpha^3 + \frac{1}{\alpha^3}) + \dots$ to 12 terms =

• A. -32
• B. 32
• C. 0
• D. 16

Hint:

The roots of $x^2-x+1=0$ are $-\omega$ and $-\omega^2$, where $\omega$ is a complex cube root of unity. Alternatively, they are $e^{i\pi/3}$ and $e^{-i\pi/3}$. Using De Moivre's theorem, $\alpha^n + 1/\alpha^n = (e^{i\pi/3})^n + (e^{-i\pi/3})^n = e^{in\pi/3} + e^{-in\pi/3} = 2\cos(n\pi/3)$. The sum becomes a sum of cosine terms, which also shows the periodic nature.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The roots of the equation $x^2-x+1=0$ are complex cube roots of -1. We can use the properties of these roots to evaluate the terms of the series. The series involves powers of $\alpha$, which will be periodic.
Step 2: Key Formula or Approach
First, identify the root $\alpha$. Multiplying the equation $x^2-x+1=0$ by $(x+1)$ gives $(x+1)(x^2-x+1)=x^3+1=0$. So, any root $\alpha$ satisfies $\alpha^3 = -1$. This implies $\alpha^6 = (\alpha^3)^2 = (-1)^2 = 1$. The powers of $\alpha$ are periodic with a period of 6. We will calculate the first few terms of the series and look for a pattern.
Step 3: Detailed Explanation
Let $S$ be the sum we want to find. The n-th term of the series is $T_n = \alpha^n + \frac{1}{\alpha^n}$. Since $\alpha$ is a root of $x^2-x+1=0$, we have $\alpha^2-\alpha+1=0$.
Dividing by $\alpha$ (since $\alpha \neq 0$), we get $\alpha - 1 + \frac{1}{\alpha} = 0$, which means $\alpha + \frac{1}{\alpha} = 1$.

Let's calculate the first 6 terms:
$T_1 = \alpha + \frac{1}{\alpha} = 1$.
$T_2 = \alpha^2 + \frac{1}{\alpha^2} = \left(\alpha + \frac{1}{\alpha}\right)^2 - 2 = 1^2 - 2 = -1$.
$T_3 = \alpha^3 + \frac{1}{\alpha^3}$. Since $\alpha^3 = -1$, this is $(-1) + \frac{1}{-1} = -1 - 1 = -2$.
$T_4 = \alpha^4 + \frac{1}{\alpha^4} = (\alpha^3\alpha) + \frac{1}{\alpha^3\alpha} = -\alpha - \frac{1}{\alpha} = -(\alpha+\frac{1}{\alpha}) = -1$.
$T_5 = \alpha^5 + \frac{1}{\alpha^5} = (\alpha^3\alpha^2) + \frac{1}{\alpha^3\alpha^2} = -\alpha^2 - \frac{1}{\alpha^2} = -(\alpha^2+\frac{1}{\alpha^2}) = -(-1) = 1$.
$T_6 = \alpha^6 + \frac{1}{\alpha^6}$. Since $\alpha^6 = 1$, this is $1 + \frac{1}{1} = 2$.
The sum of the first 6 terms is:
$S_6 = T_1 + T_2 + T_3 + T_4 + T_5 + T_6 = 1 + (-1) + (-2) + (-1) + 1 + 2 = 0$.
Since $\alpha^6 = 1$, the sequence of terms is periodic with period 6.
$T_{n+6} = \alpha^{n+6} + \frac{1}{\alpha^{n+6}} = \alpha^n \alpha^6 + \frac{1}{\alpha^n \alpha^6} = \alpha^n(1) + \frac{1}{\alpha^n(1)} = T_n$.
The sum required is for 12 terms, which is two full periods.
$S_{12} = (T_1 + \dots + T_6) + (T_7 + \dots + T_{12})$
$S_{12} = S_6 + S_6 = 0 + 0 = 0$.
Step 4: Final Answer
The sum of the series to 12 terms is 0.