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

Check the Answer


But try the problem first…

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

Source
Suggested Reading

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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