A probability problem by Santiago Hincapie

Probability Level 4

∑ n = 1 ∞ F n x n = x 505 \large{ \sum _{ n=1 }^{ \infty }{ \frac { F_{ n } }{ { x }^{ n } } =\frac { x }{ 505 } } }

Find the value of x , x > 0 x, \ x > 0 such that the above equation satisfies where F n F_n denotes the n th n^{\text{th}} Fibonacci number .


The answer is 23.

Santiago Hincapie by Santiago Hincapie , 25, Belgium

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2 solutions

Tom Van Lier
Nov 11, 2015

Using the following property :

∑ n = 1 ∞ F n x n = x x 2 − x − 1 , \sum\limits_{n=1}^\infty \dfrac{F_{n}}{x^n} = \dfrac{x}{x^2 - x - 1} ,

we deduce that x 2 − x − 1 = 505 x^2 - x - 1 = 505 , which gives us the answers -22 and 23.

⇒ 23 \Rightarrow 23 is the only positive x.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Can you proof this property?

Santiago Hincapie - 5 years, 7 months ago

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Let S = ∑ n = 1 ∞ F n x n \displaystyle S = \sum_{n=1}^\infty \frac{F_n}{x^n} . F 1 = F 2 = 1 F_1 = F_2 = 1 and F n = F n − 1 + F n − 2 F_n = F_{n-1} + F_{n-2} so

S = ∑ n = 1 ∞ F n x n = 1 x + 1 x 2 + ∑ n = 3 ∞ F n x n = 1 x + 1 x 2 + ∑ n = 3 ∞ F n − 1 + F n − 2 x n = 1 x + 1 x 2 + ∑ n = 3 ∞ F n − 1 x n + ∑ n = 3 ∞ F n − 2 x n = 1 x + 1 x 2 + 1 x ∑ n = 2 ∞ F n x n + 1 x 2 ∑ n = 1 ∞ F n x n = 1 x + [ 1 x + 1 x 2 ] ∑ n = 1 ∞ F n x n = 1 x + [ 1 x + 1 x 2 ] S \begin{aligned} S &= \sum_{n=1}^\infty \frac{F_n}{x^n} \\ &= \frac{1}{x} + \frac{1}{x^2} + \sum_{n=3}^\infty \frac{F_n}{x^n} \\ &= \frac{1}{x} + \frac{1}{x^2} + \sum_{n=3}^\infty \frac{F_{n-1}+F_{n-2}}{x^n} \\ &= \frac{1}{x} + \frac{1}{x^2} + \sum_{n=3}^\infty \frac{F_{n-1}}{x^n} + \sum_{n=3}^\infty \frac{F_{n-2}}{x^n} \\ &= \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x} \sum_{n=2}^\infty \frac{F_n}{x^n} + \frac{1}{x^2} \sum_{n=1}^\infty \frac{F_n}{x^n} \\ &= \frac{1}{x} + \left[ \frac{1}{x} + \frac{1}{x^2} \right] \sum_{n=1}^\infty \frac{F_n}{x^n} \\ &= \frac{1}{x} + \left[ \frac{1}{x} + \frac{1}{x^2} \right] S \end{aligned}

Thus, S = x x 2 − x − 1 S = \dfrac{x}{x^2-x-1} .

Jake Lai - 5 years, 6 months ago

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Turns out f ( x ) = x x 2 − x − 1 = f ( 1 x ) f(x)=\frac{x}{x^2-x-1}=f(\frac{1}{x})

Julian Poon - 5 years, 6 months ago

Thanks for writing that out, I didn't have time for it.

Tom Van Lier - 5 years, 6 months ago

did it the same way!

Divyansh Choudhary - 5 years, 5 months ago

Apply this .

Pi Han Goh - 5 years, 6 months ago
Otto Bretscher
Sep 29, 2015

I believe there is a second answer besides 23, namely, x = − 22 x=-22

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Thank you. The problem statement has been edited.

Brilliant Mathematics Staff - 5 years, 6 months ago

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