First we prove a proposition made by Fibonacci that states the difference between two squared consecutive numbers is the sum of the two numbers. For example, 5

^{2 }- 4

^{2}= 9 = 5 + 4. The proof is pretty simple, but it's good for a warmup.

Let

*n*be any integer. We will prove that (

*n*+ 1)

^{2 }-

*n*

^{2}= (

*n +*1) +

*n*directly. Distributing the lefthand side of this equation, we get

(

*n*+ 1)^{2 }-*n*^{2}=*n*^{2}+ 2*n*+ 1 -*n*^{2}.
Simplifying gives

(

*n*+ 1)^{2 }-*n*^{2}= 2*n*+ 1 = (*n*+ 1) +*n,*

which was to be shown. So, the difference between two squared consecutive numbers is the sum of the two numbers. Next, we prove a formula for the sum of the Fibonacci numbers from 1 to

*n*with an inductive proof.
Using wikipedia, I found that the sum of all Fibonacci numbers up to

*n*is one less than the (*n*+ 2)th term. That is, if*F*(*n*) is the*n*th Fibonacci number,
∑

*k*=1nF*k*= F n+2 −1.*F*(1) +

*F*(2) = 1 + 1 = 2 = 3 - 1 =

*F*(4) - 1. With the basis for induction established, we now assume for the inductive hypothesis that

∑

*i*=1kF*i*= F k+2 −1and we want show

∑

*i*=1k+1F*i*= F k+3 −1.Notice that we can rewrite this last equation as

∑

*i*=1k+1F*i*= ∑ i=1 k F i+Fk+1.The sum on the righthand side of this is our induction hypothesis, so we have

∑

*i*=1k+1F*i*= F k+2 −1 + Fk+1.*Fk*+ 1 +

*Fk*+ 2 =

*Fk*+ 3. Thus, the previous equation can be written

∑

*i*=1k+1F*i*= F k+3 −1,which completes the proof. Thus, we have shown that the sum of the Fibonacci numbers to the

*n*th term is one less than the (

*n*+ 2)th term. Many of these inductive proofs involving sums seem to be similar, and having proved the sum of natural numbers helped in my understanding of this proof. I remember doing many proofs like this in 210, and I always thought the many patterns that came about from this sequence were interesting. My plan is to continue working with the Fibonacci sequence in my next weekly entry.

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