# Problem on Series and Sequences | SMO, 2012 | Problem 23

Try this beautiful problem from Singapore Mathematics Olympiad, 2012 based on Series and Sequences.

## Problem on Series and Sequences (SMO Test)

For each positive integer $n \geq 1$ , we define the recursive relation given by $a_{n+1} = \frac {1}{1+a_{n}}$.

Suppose that $a_{1} = a_{2012}$.Find the sum of the squares of all

possible values of $a_{1}$.

• 2
• 3
• 6
• 12

### Key Concepts

Series and Sequence

Functional Equation

Recursive Relation

Challenges and Thrills - Pre - College Mathematics

## Try with Hints

If you got stuck start from here :

At first we have to understand the sequence it is following

Given that : $a_{n+1} = \frac {1}{1+a_{n}}$

Let $a_{1} = a$

so , $a_{2} = \frac {1}{1 + a_{1}}$ = $\frac {1}{1 + a}$

Again, $a_{3} = \frac {1}{1+a_{2}} = \frac {1+a}{2+a}$

For , $a_{4} = \frac {1}{1+a_{3}} = \frac {2+a}{3+2a}$

And , $a_{5} = \frac {1}{1+a_{4}} = \frac {3+2a}{5+3a}$

and so on........

Try to do the rest ..................................

Looking at the previous hint ..................

In general we can say .................

$a_{n} = \frac {F_{n} + F_{n-1}a}{F_{n+1} +F_{n}a}$

Where $F_{1} = 0 , F_{ 2} = 1$ and $F_{n+1} = F_{n}$ for all value of $n\geq 1$

Try to do the rest .......

Here is the rest of the solution,

If $a_{2012} = \frac {F_{2012}+F_{2011}a}{F_{2013} + F{2012}a} = a$

Then $(a^2+a-1 )F_{2012} = 0$

Since $F_{2012}>0$ we have $a^2 +a -1 = 0$ ............................(1)

Assume x and y are the two roots of the $eq^n (1)$, then

$x^2 + y^2 = (x+y)^2 -2xy = (-1)^2 - 2(-1) = 3$ (Answer)

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