Fibonacci

It is given that f ( 0 ) = 0 f(0)=0 , f ( 1 ) = 1 f(1)=1 , and f ( n ) = f ( n 1 ) + f ( n 2 ) f(n)=f(n-1)+f(n-2) for n 2 n \ge 2 .

What is the value of f ( 50 ) f(50) ?


The answer is 12586269025.

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1 solution

f ( n ) = F n f(n) = F_n is the Fibonacci number. We can solve for f ( 50 ) = F 50 f(50)=F_{50} using the linear recurrence relation f ( n ) = f ( n 1 ) + f ( n 2 ) f(n) = f(n-1) + f(n-2) . The characteristic polynomial of the recurrence is as follows.

r 2 r 1 = 0 r = 1 ± 5 2 where 1 + 5 2 = φ is the golden ratio. F n = c 1 φ n + c 2 ψ n where c 1 , c 2 are constants, ψ = 1 5 2 . F 0 = c 1 + c 2 = 0 c 2 = c 1 F 1 = c 1 ( φ ψ ) = 1 c 1 = 1 5 F n = φ n ψ n 5 Binet’s formula F 50 = φ 50 ψ 50 5 = 12586269025 \begin{aligned} r^2 - r - 1 & = 0 \\ \implies r & = \frac {1 \pm \sqrt 5}2 & \small \color{#3D99F6} \text{where }\frac {1+\sqrt 5}2 = \varphi \text{ is the golden ratio.} \\ \implies F_n & = c_1 \varphi^n + c_2 \color{#3D99F6} \psi^n & \small \color{#3D99F6} \text{where }c_1, \ c_2 \text{ are constants, }\psi = \frac {1-\sqrt 5}2. \\ F_0 & = c_1 + c_2 = 0 & \small \color{#3D99F6} \implies c_2 = - c_1 \\ F_1 & = c_1(\varphi -\psi) = 1 \\ \implies c_1 & = \frac 1{\sqrt 5} \\ \implies F_n & = \frac {\varphi^n - \psi^n}{\sqrt 5} & \small \color{#3D99F6} \text{Binet's formula} \\ F_{50} & = \frac {\varphi^{50} - \psi^{50}}{\sqrt 5} \\ & = \boxed{12586269025} \end{aligned}

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