Recursive Recursion - 1

Probability Level 5

$\large \begin{array} {l l } a_{n} & = a_{n+1}a_{n-1} - b_{n+1}b_{n-1} \\ b_{n} & = b_{n+1}a_{n-1} + a_{n+1}b_{n-1} \\ \end{array}$

Consider, two recursive relations defined as above with initial starting conditions $a_{0} = 1$ , $b_{0} = 2$ , $a_{1} = 3$ and $b_{1} = 4$ .

If the value of $a_{2015}^{2} + b_{2015}^{2}$ can be expressed as $\dfrac{x}{y}$ , where $x$ , $y$ are coprime integers. Find the value of $x+y$ .

The answer is 6.

1 solution

Surya Prakash
Jul 29, 2015

Consider,

$a_{n} + i b_{n} = a_{n+1}a_{n-1} + i^{2} b_{n+1}b_{n-1} + ia_{n+1}b_{n-1} + ib_{n+1}a_{n-1}$ $a_{n} + i b_{n} = (a_{n+1} + i b_{n+1})(a_{n-1} + i b_{n-1} )$

Let $S_{n} = a_{n} + i b_{n}$

So,

$S_{n}=S_{n+1}S_{n-1}=S_{n+1}S_{n}S_{n-2}$ $S_{n+1} = \dfrac{1}{S_{n-2}}$ $S_{n}= \dfrac{1}{S_{n-3}} = S_{n-6}$

So,

$S_{2015} = S_{5}=\dfrac{1}{S_{2}}$

But since $S_{1} = S_{2}S_{0}$

So,

$S_{2}=\dfrac{S_{1}}{S_{0}} =\dfrac{3+4i}{1+2i}$ $a_{2015}^{2} + b_{2015}^{2} = |S_{2015}|^{2} = \frac{1}{|S_{2}|^{2}} = \frac{1}{5}$

So, $x+y=\boxed{6}$ .

Moderator note:

Great observation with the use of the complex number substitution!

Brilliant! Do you have any resources that have more problems of this type?

Alan Yan - 5 years, 8 months ago

No. This is own problem

Surya Prakash - 5 years, 8 months ago

Recursive Recursion - 1

The answer is 6.

1 solution

Moderator note:

0 pending reports