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: Identify Roots The roots of \( x^2 - x + 1 = 0 \) are \( -\omega \) and \( -\omega^2 \), where \( \omega \) is the cube root of unity. Alternatively, \( \alpha = e^{i\pi/3} \) or \( e^{-i\pi/3} \). Note that \( \alpha^3 = -1 \) and \( \alpha^6 = 1 \).
Step 2: Evaluate General Term Let \( T_n = \left(\alpha^n + \frac{1}{\alpha^n}\right)^3 \). Since \( \alpha = \cos(\pi/3) + i\sin(\pi/3) \), \( \alpha^n + \alpha^{-n} = 2\cos(n\pi/3) \). Calculate terms for \( n=1 \) to \( 6 \):
\( n=1: 2\cos(\pi/3) = 1 \implies T_1 = 1^3 = 1 \)
\( n=2: 2\cos(2\pi/3) = -1 \implies T_2 = (-1)^3 = -1 \)
\( n=3: 2\cos(\pi) = -2 \implies T_3 = (-2)^3 = -8 \)
\( n=4: 2\cos(4\pi/3) = -1 \implies T_4 = (-1)^3 = -1 \)
\( n=5: 2\cos(5\pi/3) = 1 \implies T_5 = 1^3 = 1 \)
\( n=6: 2\cos(2\pi) = 2 \implies T_6 = 2^3 = 8 \)

Step 3: Sum the Series Sum of first 6 terms: \( 1 - 1 - 8 - 1 + 1 + 8 = 0 \). Since the sequence of values of \( 2\cos(n\pi/3) \) is periodic with period 6, the sum of every 6 terms is 0. Total Sum (12 terms) = \( 2 \times (\text{Sum of 6 terms}) = 2 \times 0 = 0 \).