카테고리

Ideas

  • Mathematical induction

Claim

Let (fn)(f_n) be the sequence of Fibonacci numbers, i.e.

f1=1,f2=1,fn=fn1+fn2 for n3.f_1 = 1, f_2 = 1, f_n = f_{n-1} + f_{n-2} \text{ for } n \geq 3.

Let (sn)=nfn(s_n) = \sum_{n}^{\infty} f_n. Then the following statement is true:

sn=fn+21.s_n = f_{n+2} - 1.

Proof

For n=1n=1, s1=f1=1s_1 = f_1 = 1. And we can know it satisfies the claim, since f31=21=1f_3 - 1 = 2 - 1 = 1.

Assume that the claim is true for some n=k,kNn=k, k \in \mathbb{N}:

sk=fk+21s_k = f_{k+2} - 1

We need to show that it also holds for n=k+1n=k+1:

sk+1=fk+31s_{k+1} = f_{k+3} - 1

From the claim:

f1+f2++fk=fk+21f_1 + f_2 + \ldots + f_k = f_{k+2} - 1

Adding fk+1f_{k+1} to both sides:

f1+f2++fk+fk+1=fk+21+fk+1f_1 + f_2 + \ldots + f_k + f_{k+1} = f_{k+2} - 1 + f_{k+1}

Using the definition of Fibonacci numbers, this can be written as:

sk+1=fk+31s_{k+1} = f_{k+3} - 1

This completes the inductive step. By the principle of mathematical induction, the claim is true for all nNn \in \mathbb{N}.