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Singapore Math Olympiad

Series Problem | SMO, 2013 | Problem 27

Try this beautiful problem from Singapore Mathematics Problem, SMO, 2012 based on Series. You may use sequential hints to solve the problem.

Try this beautiful problem from Singapore Mathematics Problem, SMO, 2012 based on Series. You may use sequential hints to solve the problem.

Series Problem – (SMO Entrance)


Find the value of \(\lfloor(\frac {3+\sqrt {17}}{2})^{6} \rfloor \)

  • 2041
  • 1504
  • 1699
  • 2004

Key Concepts


Series

Algebra

Check the Answer


But try the problem first…

Answer : 2041

Source
Suggested Reading

Singapore Mathematics Olympiad

Challenges and Thrills – Pre – College Mathematics

Try with Hints


First hint

If you got stuck in this hint please do through this hint :

We can assume two integers, \(\alpha\) and \(\beta\) where ,

\(\alpha = \frac {3+\sqrt {17}}{2}\) so let \(\beta = \frac {3 – \sqrt {17}}{2}\)

Then if we try to substitute the value in any formula lets try to find the value of \(\alpha . \beta\)

and \(\alpha + \beta \)

I think you can definitely find the values………Try …………………..

Second Hint

If you are still not getting it ………………

\(\alpha .\beta = \frac {3+\sqrt {17}}{2} . \frac {3-\sqrt {17}}{2} = -2 \)

and again \(\alpha + \beta = 3\)

Lets consider the sum \(s_{n} = \alpha ^ {n} + \beta ^{n}\).

Then try to find the value of \( 3 s_{n+1} + 2 s_{n}\) …………………………..

Final Step

This one is the last hint ………..

\( 3 s_{n+1} + 2 s_{n} = (\ (alpha +\beta)(\alpha^{n+1} + \beta^ {n+1} -\alpha\beta(\alpha^{n} + \beta ^{n}\)

= \(\alpha ^ {n+2} – \beta ^ {n+2} = s_{n+2}\)

Note that \(|\beta| < 1\) .Then for even positive integer n, \(\lfloor a^{n}\rfloor = s_{n} + \lfloor – \beta^n \rfloor = s_{n} – 1\).

As \(s_{0} = 2\) and \(s_{1}=3\) we can find \(s_{6} = 2041\)

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