Question

If $x^2 + x + 1 = 0$, find the value of
$\sum_{k=1}^{15} \left(x^k + \frac{1}{x^k}\right)^4$

• A. 90
• B. 120
• C. 150
• D. 180

Hint:

Whenever you see the equation $x^2+x+1=0$ or $x^3-1=0$, immediately think of the cube roots of unity ($\omega$). Problems involving summations of powers of $\omega$ often rely on the periodic nature of these powers (repeating every 3 terms). Group the terms based on whether the power is a multiple of 3 or not.

Answer 1

The Correct Option is A

Step 1: Understanding the Question: 
The equation $x^2 + x + 1 = 0$ is a characteristic equation whose roots are the non-real cube roots of unity, $\omega$ and $\omega^2$. We need to evaluate a summation involving powers of these roots. 
Step 2: Key Formula or Approach: 
1. The roots of $x^2 + x + 1 = 0$ are $x = \omega$ and $x = \omega^2$. 
2. Properties of cube roots of unity: - $\omega^3 = 1$ - $1 + \omega + \omega^2 = 0$ - $\frac{1}{\omega} = \omega^2$ and $\frac{1}{\omega^2} = \omega$ 3. We need to evaluate the term inside the summation, $(x^k + \frac{1}{x^k})^4$, for different values of k. 
Step 3: Detailed Explanation: 
Let's choose $x = \omega$. The expression inside the summation becomes: 
\[ \left(\omega^k + \frac{1}{\omega^k}\right)^4 = (\omega^k + \omega^{-k})^4 = (\omega^k + \omega^{2k})^4 \] (Note: Since $\omega^3=1$, $\omega^{-k} = \omega^{3m-k}$ for some integer m, which is equivalent to $\omega^{2k}$ because $\omega^k \cdot \omega^{2k} = \omega^{3k} = 1$). 
Now, we evaluate the base term $\omega^k + \omega^{2k}$ based on the value of k: 
 Case 1: k is a multiple of 3. 
Let $k = 3m$, where m is an integer. 
$\omega^k = \omega^{3m} = (\omega^3)^m = 1^m = 1$. 
$\omega^{2k} = \omega^{6m} = (\omega^3)^{2m} = 1^{2m} = 1$. 
So, $\omega^k + \omega^{2k} = 1 + 1 = 2$. The term in the sum is $(2)^4 = 16$. 
 Case 2: k is not a multiple of 3. 
From the property $1 + \omega + \omega^2 = 0$, we know that for any integer k not divisible by 3, $1 + \omega^k + \omega^{2k} = 0$. This implies $\omega^k + \omega^{2k} = -1$. The term in the sum is $(-1)^4 = 1$. 
 Now we need to apply this to the summation from $k=1$ to $k=15$: 
- The values of k that are multiples of 3 are: 3, 6, 9, 12, 15. There are 5 such terms. 
- The values of k that are not multiples of 3 are the remaining terms. Total terms = 15. So, there are $15 - 5 = 10$ such terms. 
The total sum is the sum of all terms from Case 1 and Case 2: 
\[ \text{Sum} = ( \text{Number of terms from Case 1} \times \text{Value}) + (\text{Number of terms from Case 2} \times \text{Value}) \] \[ \text{Sum} = (5 \times 16) + (10 \times 1) \] \[ \text{Sum} = 80 + 10 = 90 \] Step 4: Final Answer: 
The value of the summation is 90.