AP EAMCET · Maths · Complex Number
If \(1, \omega, \omega^2\) are the cube roots of unity, then
\(1\left(2+\frac{1}{\omega}\right)\left(2+\frac{1}{\omega^2}\right)+2\left(3+\frac{1}{\omega}\right)\left(3+\frac{1}{\omega^2}\right)+3\left(4+\frac{1}{\omega}\right)\left(4+\frac{1}{\omega^2}\right)+\ldots 10 \text { terms }=\)
- A \(3080\)
- B \(3465\)
- C \(3175\)
- D \(3715\)
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 \(T_k = k(k^2+k+1) = k^3+k^2+k\)…
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