Question

If $\omega$ is a cube root of unity and $\omega \neq 1$, then $2 \cos \left( \sum_{k=1}^{10} (k-\omega)(k-\omega^2) \frac{\pi}{1350} \right)$ is equal to

• A. 1

Answer 1

Answer: (A)

1

Answer 2

We are given that $\omega$ is a non-real cube root of unity, so

$$\omega + \omega^2 = -1 \quad \text{and} \quad \omega \omega^2 = 1.$$

Thus,

$$(k-\omega)(k-\omega^2) = k^2 - k(\omega+\omega^2) + \omega \omega^2 = k^2 + k + 1.$$

The sum becomes:

$$\sum_{k=1}^{10} (k^2+k+1) = \sum_{k=1}^{10} k^2 + \sum_{k=1}^{10} k + 10.$$

We know:

$$\sum_{k=1}^{10} k^2 = 385,\quad \sum_{k=1}^{10} k = 55.$$

Thus,

$$385 + 55 + 10 = 450.$$

Then the expression inside cosine is:

$$450\, \frac{\pi}{1350} = \frac{\pi}{3}.$$

Finally,

$$2\cos\frac{\pi}{3} = 2\left(\frac{1}{2}\right) = 1.$$