IMC 2015/3

Probability Level 4

Consider the recurrence relation with initial values $F(0) = 0, F(1) = \frac{3}{2}$ and $F(n) = \frac{5}{2} F(n-1) - F(n-2).$

To 2 decimal places, evaluate $\sum_{n=0}^ \infty \frac{1}{ F(2 ^n ) } .$

The answer is 1.00.

2 solutions

Pi Han Goh
Jul 31, 2015

By induction, we can see that $F(n) = 2^n - 1/2^n$ .

Let $x_n = 2^{2^n} \Rightarrow x_{n+1} = x_n^2$ .

$\frac1{F(2^n)} = \frac{x_n}{x_n^2 - 1} = \frac{x_n+ 1}{x_n^2 - 1} - \frac1{x_n^2 - 1} = \frac1{x_n - 1} - \frac1{x_{n+1} -1}$

Sum telescope to $\frac1{x_0 - 1} = \boxed1$ .

Moderator note:

Simple standard approach. I'm surprised that this problem was so easy.

Are you sure ? What is F (2)=¿

Camal Qedirov - 5 years, 9 months ago

Oh I didn't spot my error. Thanks for notifying me.

Pi Han Goh - 5 years, 9 months ago

Xian Ng
Aug 9, 2015

I saw plus instead of minus for the longest time and got stuff wrong with semi legit math (with plus it multiplies 29/4 every increment of 2 i found out) So i wrote code instead (y) Then i found my mistake But the code was already there with the switch of a sign so i stuck with it: http://goo.gl/Q3QeKb

IMC 2015/3

The answer is 1.00.

2 solutions

Moderator note:

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