Question

If \(ω\) is a complex cube root of unity, then the value of the expression \(2(1+\frac{1}{ω})(1+\frac{1}{ω^2})+3(2+\frac{1}{ω})(2+\frac{1}{ω^2})+\ldots +(n+1)(n+\frac{1}{ω})(n+\frac{1}{ω^2})\) is...

• A. \([\frac{n(n+1)}{2}]^2+n\)
• B. \([\frac{n(n+1)}{2}]^2-n\)
• C. \([\frac{n(n+1)}{2}]^2\)
• D. \([\frac{n(n-1)}{2}]^2\)

Hint:

Show (k+1/w)(k+1/w^2) = k^2 - k + 1, so the term is k^3 + 1.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The \(k\)-th term of the series is \((k+1)\left(k+\frac1\omega\right)\left(k+\frac1{\omega^2}\right)\) for \(k=1\) to \(n\). For a cube root of unity, \(\omega^3=1\) and \(1+\omega+\omega^2=0\).

Step 2: Simplify the product:
Since \(\frac1\omega=\omega^2\) and \(\frac1{\omega^2}=\omega\), their sum is \(\omega+\omega^2=-1\) and their product is \(1\). So
\[ \left(k+\tfrac1\omega\right)\left(k+\tfrac1{\omega^2}\right) = k^2 + k(-1) + 1 = k^2-k+1 \]

Step 3: Multiply by (k+1):
\[ (k+1)(k^2-k+1) = k^3+1 \]

Step 4: Sum the series:
\[ \sum_{k=1}^{n}(k^3+1) = \left[\frac{n(n+1)}{2}\right]^2 + n \]

Step 5: Choose:
This matches option (A). Option (B) has \(-n\) and option (C) leaves out \(n\), which come from dropping the +1 or subtracting it.

Final Answer:
The sum equals [n(n+1)/2]^2 + n, option (A). \[ \boxed{\left[\frac{n(n+1)}{2}\right]^2+n} \]