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I'm having trouble proving by induction that this following Fibonacci-Lucas equation

$$F_{2n+k} = F_n L_{n+k} + (-1)^n F_k \tag{*}$$

is true, given that

$$F_{2n} = F_nL_n$$

and

$$F_{2n+1} = F_nL_{n+1} + (-1)^n$$

are true.

I did the base case $k = 1$, but I can't prove the induction step for $k+1$. In particular, my textbook said I have to assume (*) is true for $k$ and prove it for $k+1$, but I cannot prove it without assuming (*) is true for $k$ AND $k-1$.

Can someone help me? This is the first time I'm posting so I'm sorry if there's anything wrong.

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1 Answer 1

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You need to use strong induction in this case, i.e., prove the base case(s) and then assume it's true for all cases less than some limit. Note, however, that the # of base cases to prove must be at least the # you will directly reference in your proof.

Using this variant in your case, you first need to prove the base cases of $k = 0$ and $k = 1$, which are true using what you're allowed to assume about $F_{2n}$ and $F_{2n+1}$, plus given that $F_0 = 0$ and $F_1 = 1$. Thus, assume that (*) is true for all $k$ from $0$ to some integer $m \ge 1$. Now, you can try to prove (*) for $k = m + 1$ given it's true for $k = m$ and $k = m - 1$. Since your question indicates that this is the issue which was preventing you from completing the induction, I assume you can do the rest yourself & won't give a solution to it here.

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  • $\begingroup$ Thank you for the reply! So basically I just need to do induction as usual, but instead of assuming ( *) is only true for k I must assume it's true for 0,1,2... k. Is that correct? Is there any way to prove ( *) by normal induction instead of strong induction? $\endgroup$
    – chocolatte
    Commented Apr 21, 2019 at 0:01
  • $\begingroup$ @HangNguyen You're welcome. Yes, as I stated, you just need to assume it's true from $0$ to some $m \ge 1$. The problem with trying to prove (*) using normal induction instead is you are quite limited if you can only assume the previous value. Also, since both Fibonacci and Lucas definitions involve the sum of the previous $2$ values, you likely can't use them but, instead, must use the direct definitions of what the $F_n$ & $L_n$ values. However, not only will this be more difficult, but you won't even need induction then. Nonetheless, you may wish to check it further anyway to verify this. $\endgroup$ Commented Apr 21, 2019 at 0:21

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