` sin "" (pi)/( 900) { sum_(r = 1)^(10) ( r - omega ) ( r - omega ^(2))}` =

Options

A. `- 1 `
B. 0
C. 1
D. `sqrt(3) // 2 `

Answer

To solve the given problem, we need to evaluate the expression:

\[ \sin\left(\frac{\pi}{900} \sum_{r=1}^{10} (r - \omega)(r - \omega^2)\right) \]

where \(\omega\) is a complex cube root of unity. Let's break this down step by step.

### Step 1: Expand the product \((r - \omega)(r - \omega^2)\)

Using the distributive property, we can expand the product:

\[ (r - \omega)(r - \omega^2) = r^2 - r\omega - r\omega^2 + \omega\omega^2 \]

Since \(\omega\) and \(\omega^2\) are roots of unity, we know that:

\[ \omega \cdot \omega^2 = \omega^3 = 1 \]

Thus, we can simplify the expression:

\[ (r - \omega)(r - \omega^2) = r^2 - r(\omega + \omega^2) + 1 \]

### Step 2: Use the property of cube roots of unity

From the properties of cube roots of unity, we know:

\[ 1 + \omega + \omega^2 = 0 \implies \omega + \omega^2 = -1 \]

Substituting this back into our expression gives:

\[ (r - \omega)(r - \omega^2) = r^2 + r + 1 \]

### Step 3: Sum the expression from \(r = 1\) to \(10\)

Now we need to compute the sum:

\[ \sum_{r=1}^{10} (r^2 + r + 1) \]

This can be broken down into three separate sums:

\[ \sum_{r=1}^{10} r^2 + \sum_{r=1}^{10} r + \sum_{r=1}^{10} 1 \]

### Step 4: Calculate each sum

1. **Sum of squares**: The formula for the sum of squares of the first \(n\) natural numbers is:

\[ \sum_{r=1}^{n} r^2 = \frac{n(n + 1)(2n + 1)}{6} \]

For \(n = 10\):

\[ \sum_{r=1}^{10} r^2 = \frac{10 \cdot 11 \cdot 21}{6} = 385 \]

2. **Sum of first \(n\) natural numbers**: The formula for the sum of the first \(n\) natural numbers is:

\[ \sum_{r=1}^{n} r = \frac{n(n + 1)}{2} \]

For \(n = 10\):

\[ \sum_{r=1}^{10} r = \frac{10 \cdot 11}{2} = 55 \]

3. **Sum of \(1\)**: Since we are summing \(1\) for \(10\) terms:

\[ \sum_{r=1}^{10} 1 = 10 \]

### Step 5: Combine the results

Now, we can combine these results:

\[ \sum_{r=1}^{10} (r^2 + r + 1) = 385 + 55 + 10 = 450 \]

### Step 6: Substitute back into the sine function

Now we substitute back into the sine function:

\[ \sin\left(\frac{\pi}{900} \cdot 450\right) = \sin\left(\frac{450\pi}{900}\right) = \sin\left(\frac{\pi}{2}\right) \]

### Step 7: Evaluate the sine function

We know that:

\[ \sin\left(\frac{\pi}{2}\right) = 1 \]

### Final Answer

Thus, the final answer is:

\[ \sin\left(\frac{\pi}{900} \sum_{r=1}^{10} (r - \omega)(r - \omega^2)\right) = 1 \]