Tag: series

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 \(a_{1}\).

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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