Try this beautiful problem from Singapore Mathematics Problem, SMO, 2012 based on Series. You may use sequential hints to solve the problem.
Find the value of \(\lfloor(\frac {3+\sqrt {17}}{2})^{6} \rfloor \)
Series
Algebra
Answer : 2041
Singapore Mathematics Olympiad
Challenges and Thrills - Pre - College Mathematics
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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