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The Fibonacci sequence $(0, 1, 1, 2, 3, 5, 8, 13, ...)$ is defined by $F_0 = 0, F_1 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for all n >= 2.

Prove that for all positive integers n that $F_1^2 + F_2^2 + ... + F_n^2 = F_nF_{n+1}$

My question concerns the base case of this induction. It says to prove this "statement" / equation for all positive integers; but, shouldn't the base case start at $n = 2$?

I'm looking for clarification at this step; hence, the proof-explanation tag.

Edit: Regarding the possible duplicate; my question is just focusing on whether I have potentially two base cases. I'm pretty sure I could be overthinking the base case (due to seeing that expression being true "for all n >= 2". As for actually finishing the proof; I'll be fine with that.

Doctor Questions
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  • "Prove that for all positive integers (including $1$) that..." so your base case starts at $n=1$. After all, it asks you to prove it for $n=1$ as well so you can't just skip over it. – JMoravitz Aug 12 '17 at 19:10
  • Actually, I didn't see anything wrong the indices or subscripts; however, I did add in the rest of the LHS of the equation. Quickly looking at that other link; it starts at n = 1. I was starting to think if there might be two base cases for this question. Base case just seems very trivial for the LHS of this equation. – Doctor Questions Aug 12 '17 at 19:13
  • You still have written $F_n=F_{n-1}+\color{red}{F_{n-1}}=2F_{n-1}\neq F_{n-1}+F_{n-2}$. The statement is true for all natural $n\geq 0$ including zero as well (since the LHS would be the emptysum hence zero and the right hand side would be $F_0\cdot F_1=0\cdot1=0$). You can use multiple base cases if you like (and in some problems it will be necessary), but it is not necessary here (if you are proving it just for positive integers). You can use the $n=1$ case as your only base case and be done with it. (if proving for all naturals including zero, you will need $n=0$ and $n=1$) – JMoravitz Aug 12 '17 at 19:16
  • The only thing which could possibly cause trouble by using a base case too small here is if you pick a base case too small (such as $n=0$ as your base case) that you might not be able to rely on the recursive nature of the fibonacci sequence which starts at $n=2$. But, by using $n=1$ as your base case, for all $k>n=1$ (i.e. for all $k\geq 2$) you can use $F_k=F_{k-1}+F_{k-2}$ succesfully. (note that $F_1=F_0+F_{-1}$ doesn't work with how you've defined the fibonacci sequence). The recursive nature of the sequence will play an important role in the inductive step. – JMoravitz Aug 12 '17 at 19:19
  • Yeah, that should be n-2. I thought there was a problem being noted with my formatting of the indices and subscripts. Given that the question is avoiding natural numbers (including zero) and just asks to prove for positive integers; the writer probably wanted to avoid issues like that when introducing mathematical induction (with a question like this). I will just stick to n = 1. As for noting that $F_1 = F_0 + F_{-1}$ won't work; this is true. The writer did note that is works for all n >= 2, though. – Doctor Questions Aug 12 '17 at 19:44
  • What may be slightly confusing is that the $n$ is used in two statements - first in the definition of $F_n$, for which we need $n\ge2$, and second in the statement of the problem, in which $n$ stands for all positive integers. These are not the same $n$. In both these statements $n$ is a kind of self-contained placeholder or label. You can rename $n$ throughout either one of the statements to for example $m$ without changing the meaning. Doing so does avoid the reuse of the label $n$, disambiguating things a bit. – Jaap Scherphuis Aug 12 '17 at 19:47
  • @Jaap, that sounds about right. $F_0 = 0$ and $F_1 = 1$ are defined before defining $F_n$. This would be to avoid that the definition of $F_n$ doesn't work for $F_1$ (the definition for $F_n$ starts at n = 2, here, anyway). At the same time; this is why I thought it seemed trivial to just put something into the LHS of the equation (for the base case n = 1) without having to derive anything. – Doctor Questions Aug 12 '17 at 19:56
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    Base cases are often trivial because they generally involve very small numbers for which the statement obviously holds true, and rarely involve the need to derive anything. In this case $F_1^2=F_1F_2$ works fine as a base case (with $n=1$) and is obviously true since $F_1=F_2=1$. – Jaap Scherphuis Aug 12 '17 at 20:07
  • A trivial mechanical proof using telescopy: simply verify $,g_n = F_n F_{n+1},$ satsifies $,g_{n+1}-g_n = F_{n+1}^2$ $\qquad$ – Bill Dubuque Aug 12 '17 at 20:17
  • It's a little pedantic but this Question is concerned (despite some initial missteps) with picking the appropriate base case (not the induction step), and the proposed duplicate might not satisfy the OP here. – hardmath Aug 12 '17 at 21:10
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    Dont think its been answered, but what is wrong with base case $n=1$? If you let $n=1$ then you have $F_1^2 = F_1 F_2$, which works out numerically to $1=1$. If you just plugged it in, you would have seen that $n=1$ works perfectly fine as a base case. I really dont understand where the confusion is coming from, nor why so many people are contributing their two cents when there is no issue in the first place. The problem is that you and others seem to be jumping to conclusions without doing any of the math. – CogitoErgoCogitoSum Aug 12 '17 at 21:47

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