Series and Trigonometry | ISI B.Stat Entrance 2009

We are going to discuss about Series and Trigonometry from I.S.I. B.Stat Entrance Objective Problem (2009).

Given that $k(1+2+3++...+n)$= $(1^2+2^2+...+n^2)$ find $cos^{-1}\frac{2n-3k}{2}$.

$(1^2+2^2+...+n^2)=\frac{n(n+1)(2n+1)}{6}$

$(1+2+...+n)=\frac{n(n+1)}{2}$

$\frac{kn(n+1)}{2}=\frac{n(n+1)(2n+1)}{6}$

then k=$\frac{2n+1}{3}$

$cos^{-1}(\frac{2n-3(\frac{2n+1}{3})}{2})$

$=cos^{-1}(\frac{-1}{2})$

$=\frac{4\pi}{3}, \frac{2\pi}{3}$

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