Question

Let \(\omega\) is an imaginary cube root of unity, then the value of \(2(1+\omega)(1+\omega^2) + 3(2\omega+1)(2\omega^2+1) + \cdots + (n+1)(n\omega+1)(n\omega^2+1)\) is

• A. \(\left[\frac{n(n+1)}{2}\right]^2 + n\)
• B. \(\left[\frac{n(n+1)}{2}\right]^2\)
• C. \(\left[\frac{n(n+1)}{2}\right]^2 - n\)
• D. None of these

Hint:

\(1 + \omega + \omega^2 = 0\), \(\omega^3 = 1\), \(\omega^2 = \bar{\omega}\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Use \(1 + \omega + \omega^2 = 0\) and \(\omega^3 = 1\).
Step 2: Detailed Explanation:
\((k\omega + 1)(k\omega^2 + 1) = k^2\omega^3 + k(\omega + \omega^2) + 1 = k^2 - k + 1\)
Sum = \(\sum_{k=1}^n (k+1)(k^2 - k + 1) = \sum_{k=1}^n (k^3 + 1)\)
= \(\sum k^3 + \sum 1 = \left[\frac{n(n+1)}{2}\right]^2 + n\)
Step 3: Final Answer:
\(\left[\frac{n(n+1)}{2}\right]^2 + n\).