Question

If $k$ times the sum of first $n$ natural numbers is equal to the sum of the squares of the first $n$ natural numbers, then ${\cos }^{-1}\left(\frac{2n-3k}{2}\right)$ is equal to

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{3}$
• C. $\frac{2\pi }{3}$
• D. $\frac{5\pi }{6}$

Answer 1

Answer: (C)

$2n-3k=-1$