Proof by induction on the first formula

Current Project
Modified Fibonacci sequence
3-month life-span

My original intention for proving the first formula to be true turned out to only loop on itself; the proof depended on the conjecture being true in the first place. Therefore, I am going to try a different approach. I want to use mathematical induction, a method which can be described much like falling dominoes. That is, the first domino is knocked over by me, knocking over the second domino, which knocks over the third domino, and so on. In theory, I could also begin by knocking over the 10th, which will knock over the 11th, then the 12th, and so on.

Applying this same idea to my sequence, the 3MLS (three-month life-span)sequence, I should be able to begin in any month and generate the sequence forward. Knowing how to do this will allow me to change my starting point (like in the domino arrangement), which will then make mathematical induction possible.


Q: What do I need to know about my rabbits in any given month in order to determine the population for the following month?

A: I need to know how many newborn, one-month old, and two-month old rabbits are in the current population. The table below shows how many of each age group are present in the first twelve months.

As you can see, the 3MLS sequence is present in each column. For the sake of comparison, let's view a similar table pertaining to the Fibonacci sequence.

Once again, the sequence (this time the Fibonacci sequence) is present in each column.

Therefore, each term in both sequences can now be represented as the sum of three terms rather than two. We have Tn = Tn-3 + Tn-4 + Tn-5. Now, let's use this to prove the first formula: Tn = Tn-2 + Tn-3.


Given: Tn = Tn-3 + Tn-4 + Tn-5
Prove: Tn = Tn-2 + Tn-3

Step 1: Show true for the smallest case
T4 = T2 + T1
2 = 1 + 1
True

Step 2: Assume true for all cases from the smallest to k, that is 4<=n<=k. Tk = Tk-2 + Tk-3 assumed true

Step 3: Show true for case k+1
T(k+1) = T(k+1)-2 + T(k+1)-3
Simplify
T(k+1) = Tk-1 +Tk-2

Use what is given, so
T(k+1) = T(k+1)-3 +T(k+1)-4 +T(k+1)-5
Simplify
* T(k+1) = Tk-2 + Tk-3 + Tk-4

Because we assumed true for all cases 4 <= n <= k this includes k-1. So, we know T(k-1) = T(k-1)-2 + T(k-1)-3 T(k-1) = Tk-3 + Tk-4 Substitute this into the * equation and we have
T(k+1) = Tk-2 + Tk-1

The proof is now concluded, and Tn = Tn-2 + Tn-3 has been found to be true.