Fibonacci in 2015

Number Theory Level 5

Let $f(n)$ be the $n$ th Fibonacci number, where $f(0) = 0$ and $f(1) = 1$ . Which of the following is equal to

$f(2015^{2015} + 2020) - f(2015^{2015} + 2010)?$

2015\times f(2015^{2015})

2015!

f(2015^{2016})

11\times f(2015^{2015} + 2015)

2014!\times f(2015!)

2015\times f(2015^2) + 10

3\times f(2015^{2012})

1 solution

Caleb Townsend
Feb 5, 2015

Recall that $f(n+10) = f(n+9) + f(n+8)$ by the recursive definition of the Fibonacci sequence. Continuing to simplify, we get $f(n+9) + f(n+8) = 2 f(n+8) + f(n+7).$ If we continue this pattern, we find that $f(n+10) = 55f(n+1) + 34f(n)$ Therefore $f(n+10) - f(n) = 55f(n+1) + 33f(n).$ Using the recursive definition again, this simplifies to $f(n+10) - f(n) = 33f(n+2) + 22f(n+1).$ Repeating this process, we see that $f(n+10) - f(n) = 11f(n+5)$ Now notice that if $n = 2015^{2015} + 2010$ , then $f(2015^{2015} + 2020) - f(2015^{2015} + 2010) = f(n + 10) - f(n).$ Therefore, the answer is $11f(n+5) = \boxed{11f(2015^{2015} + 2015)}$

Fibonacci in 2015

1 solution

0 pending reports