Search any question & find its solution
Question:
Answered & Verified by Expert
If $\rho$ is a complex cube root of unity, then
$\cos \left(\sum_{k=1}^7(k-\omega)\left(k-\omega^2\right) \frac{\pi}{175}\right)=$
Options:
$\cos \left(\sum_{k=1}^7(k-\omega)\left(k-\omega^2\right) \frac{\pi}{175}\right)=$
Solution:
2294 Upvotes
Verified Answer
The correct answer is:
$-1$
We have to calculate
$\cos \left(\sum_{k=1}^7(k-\omega)\left(k-\omega^2\right) \frac{\pi}{175}\right)$
$=\cos \left(\sum_{k=1}^7\left(k^2-k \omega^2-k \omega+\omega^3\right) \frac{\pi}{175}\right)$
$=\cos \left(\sum_{k=1}^7\left(k^2-k x(-1)+1\right) \frac{\pi}{175}\right)$
$=\cos \left(\frac{\pi}{175}\left(\sum_{k=1}^7 k^2+\sum_{k=1}^7 k+\sum_{k=1}^7 1\right)\right)$
$=\cos \left(\frac{\pi}{175}\left(\frac{1}{6} \times 7 \times(7+1) \times(2 \times 7+1)+\frac{1}{2} \times 7 \times 8+7\right)\right)$
$=\cos \left(\frac{\pi}{175}(140+28+7)\right)$
$=\cos \left(\frac{\pi}{175} \times 175\right)=\cos (\pi)=-1$.
$\cos \left(\sum_{k=1}^7(k-\omega)\left(k-\omega^2\right) \frac{\pi}{175}\right)$
$=\cos \left(\sum_{k=1}^7\left(k^2-k \omega^2-k \omega+\omega^3\right) \frac{\pi}{175}\right)$
$=\cos \left(\sum_{k=1}^7\left(k^2-k x(-1)+1\right) \frac{\pi}{175}\right)$
$=\cos \left(\frac{\pi}{175}\left(\sum_{k=1}^7 k^2+\sum_{k=1}^7 k+\sum_{k=1}^7 1\right)\right)$
$=\cos \left(\frac{\pi}{175}\left(\frac{1}{6} \times 7 \times(7+1) \times(2 \times 7+1)+\frac{1}{2} \times 7 \times 8+7\right)\right)$
$=\cos \left(\frac{\pi}{175}(140+28+7)\right)$
$=\cos \left(\frac{\pi}{175} \times 175\right)=\cos (\pi)=-1$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.