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: General term:
Write $T_k=(k+1)\,(k+\omega^2)(k+\omega)$ using $1/\omega=\omega^2$.

Step 2: Expand:
$(k+\omega)(k+\omega^2)=k^2+k(\omega+\omega^2)+\omega^3 = k^2-k+1$.

Step 3: Use sum of cubes:
$(k+1)(k^2-k+1)=k^3+1$.

Step 4: Add:
$\sum k^3=(n(n+1)/2)^2$ and $\sum 1 = n$. Total $= (n(n+1)/2)^2 + n$.

Step 5: Test:
For $n=1$: the series is just one term, $2(1+\omega^2)(1+\omega)=2\cdot 1=2$. Formula gives $1+1=2$. Good.

Final Answer:
Each term reduces to k^3 + 1. \[ \boxed{A} \]