Q48 / 160 · AP EAMCET 2023 (16 May Shift 1) · Mathematics

If $\rho$ is a complex cube root of unity, then $\cos \left(\sum_{k=1}^7(k-\omega)\left(k-\omega^2\right) \frac{\pi}{175}\right)=$

  • Correct answer
    A
    $-1$
  • B
    0
  • C
    1
  • D
    5

Solution

We have to calculate $\cos \left(\sum_{k=1}^7(k-\omega)\left(k-\omega^2\right) \frac{\pi}{175}\right)$ $=\cos \left(\sum_{k=1}^7\left(k^2-k \omega^2-k \omega+\omega^3\right) \frac{\pi}{175}\right)$ $=\cos \left(\sum_{k=1}^7\left(k^2-k x(-1)+1\right) \frac{\pi}{175}\right)$ $=\cos \left(\frac{\pi}{175}\left(\sum_{k=1}^7 k^2+\sum_{k=1}^7 k+\sum_{k=1}^7 1\right)\right)$ $=\cos \left(\frac{\pi}{175}\left(\frac{1}{6} \times 7 \times(7+1) \times(2 \times 7+1)+\frac{1}{2} \times 7 \times 8+7\right)\right)$ $=\cos \left(\frac{\pi}{175}(140+28+7)\right)$ $=\cos \left(\frac{\pi}{175} \times 175\right)=\cos (\pi)=-1$.

Hence, the correct answer is option: 'A'.

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