Question

If $ \alpha $ is a root of the equation $ x^2 + x + 1 = 0 $ and $$ \sum_{k=1}^{n} \left( \alpha^k + \frac{1}{\alpha^k} \right)^2 = 20$$, then n is equal to _______

Hint:

For sums involving roots of unity, use the fact that powers of roots of unity repeat after a certain number of terms, which simplifies the sum.

Answer 1

Correct Answer: 11

We are given that \( \alpha \) is a root of the equation \( x^2 + x + 1 = 0 \), so: \[ \alpha = \omega \] where \( \omega \) is a cube root of unity. 
Therefore, \( \alpha = \omega \) and we have the identity \( \omega^3 = 1 \). 
Now, consider the given summation: \[ \sum_{k=1}^{n} \left( \alpha^k + \frac{1}{\alpha^k} \right)^2 \] Since \( \alpha = \omega \), we can write the expression as: \[ \left( \omega^k + \frac{1}{\omega^k} \right)^2 = \omega^{2k} + \omega^k + 2 \] Now, simplifying the sum: \[ \sum_{k=1}^{n} \left( \omega^{2k} + \omega^k + 2 \right) \] This simplifies to: \[ \sum_{k=1}^{n} \omega^{2k} + \sum_{k=1}^{n} \omega^k + 2n \] We know that \( \omega^3 = 1 \), so the powers of \( \omega \) repeat every 3 terms. 
Therefore, the sum can be simplified as follows. 
The sum of powers of \( \omega \) for \( n = 3m \) (where \( m \) is some integer) is 0 for the periodic terms, and we are left with: \[ 2n = 20 \quad \Rightarrow \quad n = 10 \] 
Thus, the correct answer is \( 11 \).

Answer 2

Step 1: The expression is: 

\[ \alpha = \omega \]

Step 2: Simplifying the equation:

\[ \left( \omega^k + \frac{1}{\omega^k} \right)^2 = \omega^{2k} + \frac{1}{\omega^{2k}} + 2 \] This simplifies to: \[ \omega^{2k} + \omega^k + 2 \quad \Rightarrow \quad \omega^{3k} = 1 \]

Step 3: Summing the series:

\[ \sum_{k=1}^{n} \left( \omega^{2k} + \omega^k + 2 \right) = 20 \] This simplifies to: \[ \left( \omega^2 + \omega^4 + \omega^6 + \cdots + \omega^{2n} \right) + \left( \omega + \omega^2 + \omega^3 + \cdots + \omega^n \right) + 2n = 20 \]

Step 4: Testing possible values for \( n \):

If \( n = 3m \), then:

\[ 0 + 0 + 2n = 20 \quad \Rightarrow \quad n = 10 \quad (\text{not satisfied}) \]

If \( n = 3m + 1 \), then:

\[ \omega^2 + \omega + 2n = 20 \] \[ -1 + 2n = 20 \quad \Rightarrow \quad n = \frac{21}{2} \quad (\text{not possible}) \]

If \( n = 3m + 2 \), then:

\[ \left( \omega^8 + \omega^{10} \right) + \left( \omega^4 + \omega^5 \right) + 2n = 20 \] This simplifies to: \[ \left( \omega^2 + \omega \right) + \left( \omega + \omega^2 \right) + 2n = 20 \] \[ 2n = 22 \quad \Rightarrow \quad n = 11 \]

Step 5: Final solution:

The value of \( n \) is \( 11 \), which satisfies \( n = 3m + 2 \).