Question

If \(1, \omega, \omega^2\) are the cube roots of unity, then \( \left(1\left(2+\frac{1}{\omega}\right)\left(2+\frac{1}{\omega^2}\right) + 2\left(3+\frac{1}{\omega}\right)\left(3+\frac{1}{\omega^2}\right) + 3\left(4+\frac{1}{\omega}\right)\left(4+\frac{1}{\omega^2}\right) + \dots + 10 \text{ terms}\right) = \)

• A. \( 3080 \)
• B. \( 3465 \)
• C. \( 3175 \)
• D. \( 3715 \)

Hint:

Remember the fundamental properties of cube roots of unity: \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \). These identities are crucial for simplifying expressions involving \(\omega\) and \(\omega^2\). Specifically, \( \frac{1}{\omega} = \omega^2 \) and \( \frac{1}{\omega^2} = \omega \) are very common transformations. Also, recognizing the sum as a sum of powers of integers allows direct application of standard summation formulas.

Answer 1

The Correct Option is B

We are given that \(1, \omega, \omega^2\) are the cube roots of unity. This means: 1. \( \omega^3 = 1 \) 2. \( 1 + \omega + \omega^2 = 0 \) Let's analyze the general term in the series. The \(k\)-th term (starting from \(k=1\)) looks like: \( k \left( (k+1) + \frac{1}{\omega} \right) \left( (k+1) + \frac{1}{\omega^2} \right) \) Let's simplify the product \( \left( (k+1) + \frac{1}{\omega} \right) \left( (k+1) + \frac{1}{\omega^2} \right) \). This is in the form \((A+B)(A+C)\) where \(A = (k+1)\), \(B = \frac{1}{\omega}\), \(C = \frac{1}{\omega^2}\). The product is \( (k+1)^2 + (k+1)\left(\frac{1}{\omega} + \frac{1}{\omega^2}\right) + \frac{1}{\omega \cdot \omega^2} \) We know \( \frac{1}{\omega} = \frac{\omega^2}{\omega^3} = \omega^2 \) and \( \frac{1}{\omega^2} = \frac{\omega}{\omega^3} = \omega \). Also, \( \frac{1}{\omega \cdot \omega^2} = \frac{1}{\omega^3} = \frac{1}{1} = 1 \). And \( \frac{1}{\omega} + \frac{1}{\omega^2} = \omega^2 + \omega \). Since \( 1 + \omega + \omega^2 = 0 \), we have \( \omega + \omega^2 = -1 \). Substitute these values back into the product: \( (k+1)^2 + (k+1)(-1) + 1 \) \( (k+1)^2 - (k+1) + 1 \) Let \(m = k+1\). Then the expression is \( m^2 - m + 1 \). So the \(k\)-th term of the series is \( k ( (k+1)^2 - (k+1) + 1 ) \). Let \( T_k = k( (k+1)^2 - (k+1) + 1 ) \). The series has 10 terms, so we need to find \( \sum_{k=1}^{10} T_k \). For \(k=1\), the term is \( 1 \left( (1+1)^2 - (1+1) + 1 \right) = 1 (2^2 - 2 + 1) = 1(4 - 2 + 1) = 3 \). For \(k=2\), the term is \( 2 \left( (2+1)^2 - (2+1) + 1 \right) = 2 (3^2 - 3 + 1) = 2(9 - 3 + 1) = 2(7) = 14 \). For \(k=3\), the term is \( 3 \left( (3+1)^2 - (3+1) + 1 \right) = 3 (4^2 - 4 + 1) = 3(16 - 4 + 1) = 3(13) = 39 \). The sum is \( S = \sum_{k=1}^{10} k((k+1)^2 - (k+1) + 1) \) \( S = \sum_{k=1}^{10} k(k^2 + 2k + 1 - k - 1 + 1) \) \( S = \sum_{k=1}^{10} k(k^2 + k + 1) \) \( S = \sum_{k=1}^{10} (k^3 + k^2 + k) \) We use the sum formulas: \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \) \( \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \) \( \sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2 \) For \( n=10 \): \( \sum_{k=1}^{10} k = \frac{10(10+1)}{2} = \frac{10 \times 11}{2} = 5 \times 11 = 55 \) \( \sum_{k=1}^{10} k^2 = \frac{10(10+1)(2 \times 10+1)}{6} = \frac{10 \times 11 \times 21}{6} = \frac{2310}{6} = 385 \) \( \sum_{k=1}^{10} k^3 = \left(\frac{10(10+1)}{2}\right)^2 = (55)^2 = 3025 \) Now, sum them up: \( S = 3025 + 385 + 55 \) \( S = 3410 + 55 \) \( S = 3465 \)