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AP EAMCET · Maths · Complex Number

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

  1. A \(3080\)
  2. B \(3465\)
  3. C \(3175\)
  4. D \(3715\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(3465\)

Step-by-step Solution

Detailed explanation

\(\left((k+1)+\frac{1}{\omega}\right)\left((k+1)+\frac{1}{\omega^2}\right) = ((k+1)+\omega^2)((k+1)+\omega) = (k+1)^2+(k+1)(\omega+\omega^2)+\omega^3 = (k+1)^2-(k+1)+1 = k^2+k+1\) General term \(T_k = k(k^2+k+1) = k^3+k^2+k\)…