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# Series and Trigonometry | ISI B.Stat Entrance 2009

## Problem - Series and Trigonometry (ISI B.Stat Entrance)

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 $cos^{-1}\frac{2n-3k}{2}$.

• $\frac{\pi}{3}$
• $\frac{\pi}{2}$
• $\frac{\pi}{6}$
• $\frac{4\pi}{3}$, $\frac{2\pi}{3}$

### Key Concepts

Series

Trigonometry

Angles

Answer: $\frac{4\pi}{3}$, $\frac{2\pi}{3}$

I.S.I. B.Stat Entrance Objective Problem (2009)

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

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

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

Second Hint

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

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

Final Step

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

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

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

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