Question

$\sin \left(\frac{\pi }{900}\left\{\sum _{r=1}^{10}(r-\omega )(r-{\omega }^2)\right\}\right)=$

• A. $-1$
• B. $0$
• C. $1$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

Answer: (C)

$1$

$(r-w)(r-w^2)$
$=r^2-(w+w^2)r+w^3$
$=r^2+r+1$
Hence
$\sum (r-w)(r-w^2)$

$=\sum r^2+r+1$

$=\sum r^2+\sum r+\sum 1$

$={[\frac{r(r+1)(2r+1)}{6}+\frac{r(r+1)}{2}+r]|}_{r=10}$

$=\frac{10\left(11\right)\left(21\right)}{6}+\frac{10\left(11\right)}{2}+10$

$=385+55+10$
$=450$

Hence
$\sin \left(450.\frac{\pi }{900}\right)$
$=\sin \left(\frac{\pi }{2}\right)$
$=1$

Hence, option 'C' is correct.