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, 52 - 42 = 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 - n2 = (n + 1) + n directly. Distributing the lefthand side of this equation, we get
(n + 1)2 - n2 = n2 + 2n + 1 - n2.
Simplifying gives
(n + 1)2 - n2 = 2n + 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 nth Fibonacci number,
∑k =1nFk= F n+2 −1.
∑i =1kFi= F k+2 −1
and we want show
∑i=1k+1Fi= F k+3 −1.
Notice that we can rewrite this last equation as
∑i=1k+1Fi= ∑ i=1 k F i+Fk+1.
The sum on the righthand side of this is our induction hypothesis, so we have
∑i=1k+1Fi= F k+2 −1 + Fk+1.
∑i=1k+1Fi= F k+3 −1,
which completes the proof. Thus, we have shown that the sum of the Fibonacci numbers to the nth 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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