Try this beautiful problem from Singapore Mathematics Olympiad, 2012 based on Series and Sequences.
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 \(a_{1}\).
Series and Sequence
Functional Equation
Recursive Relation
Answer: 3
Singapore Mathematics Olympiad
Challenges and Thrills - Pre - College Mathematics
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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