Series Problem | SMO, 2013 | Problem 27

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

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 .......................

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 \( 3 s_{n+1} + 2 s_{n}\) ................................

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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