Question

\[ \sum_{k=1}^{n} \left( \alpha^k + \frac{1}{\alpha^k} \right)^2 = 20, \quad \alpha \text{ is one of the roots of } x^2 + x + 1 = 0, \text{ then } n = ? \]

• A. 3
• B. 4
• C. 5
• D. 6

Hint:

The roots of the equation \( x^2 + x + 1 = 0 \) are the cube roots of unity. Using properties of these roots simplifies the sum of the given expression.

Answer 1

The Correct Option is C

We are given the summation: \[ \sum_{k=1}^{n} \left( \alpha^k + \frac{1}{\alpha^k} \right)^2 = 20 \] where \( \alpha \) is one of the roots of the quadratic equation: \[ x^2 + x + 1 = 0 \] The roots of the quadratic equation are the cube roots of unity, given by: \[ \alpha = e^{i\frac{2\pi}{3}} \quad \text{or} \quad \alpha = e^{i\frac{4\pi}{3}} \] For both of these roots, we know that: \[ \alpha^3 = 1 \quad \text{and} \quad \alpha + \frac{1}{\alpha} = -1 \] Thus, we can simplify the term \( \left( \alpha^k + \frac{1}{\alpha^k} \right)^2 \): \[ \left( \alpha^k + \frac{1}{\alpha^k} \right) = 2\cos\left( \frac{2k\pi}{3} \right) \] So, \[ \left( \alpha^k + \frac{1}{\alpha^k} \right)^2 = 4\cos^2\left( \frac{2k\pi}{3} \right) \] Now, we sum this for \( k = 1, 2, \dots, n \). We know that for the cube roots of unity, the sum of the squares of \( \left( \alpha^k + \frac{1}{\alpha^k} \right) \) over three terms gives 3 (since each term contributes 1). Thus, to get a total sum of 20, we must have \( n = 5 \), as: \[ 5 \times 4 = 20 \] Therefore, the correct value of \( n \) is \( \boxed{5} \). Thus, the correct answer is (3) 5.