 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

But try the problem first…

Source

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 $s_{6} = 2041$