How to prove this Fibonacci sequence statement: $F^2_{n+1} - F^2_{n-1} = F_{2n}$

Question

So the statement is:

Prove: $$F^2_{n+1} - F^2_{n-1} = F_{2n}$$

How could this be proven with induction?

What about without induction?

For induction I got here: (already tested $n=1$, and assumed $n=k$ held true)

$$F^2_{k+2} - F^2_{k} = F_{2k + 2}$$ $$(F_{k+1} + F_{k})^2 - (F_{k-1} + F_{k-2})^2 = F_{2k+1} + F_{2k}$$ Then you get some cancelation based off the $n=k$ case $$2F_{k+1}F_k + F_k^2 - 2F_{k-1}F_{k-2} - F_{k-2}^2 = F_{2k+1}$$

This is where I am stuck for the induction proof.

I have no idea where to begin with the non-induction proof.

could you show me the simplification? I don't understand where to go — mathkid225, Mar 13 '18 at 23:45
https://math.stackexchange.com/questions/65011/prove-this-formula-for-the-fibonacci-sequence — lab bhattacharjee, Mar 14 '18 at 00:33
See also: Induction Proof for $F_{2n} = F^2_{n+1} - F^2_{n-1}$ — Martin Sleziak, Nov 07 '20 at 11:52

Rócherz · Accepted Answer · 2018-03-14T02:09:37.893

6

Consider this matrix: $$A=\begin{bmatrix}1&1\\1&0\end{bmatrix}.$$ Matrix $A$ is such that $$A^n=\begin{bmatrix}F_{n+1}&F_{n}\\F_{n}&F_{n-1}\end{bmatrix}.$$ Compute both $(A^n)^2$ and $A^{(2n)}$ (they are equal) and compare them entrywise: $$\begin{align}(A^n)^2&= \begin{bmatrix}F_{n+1}&F_{n}\\F_{n}&F_{n-1}\end{bmatrix} \cdot\begin{bmatrix}F_{n+1}&F_{n}\\F_{n}&F_{n-1}\end{bmatrix} \\ &=\begin{bmatrix}F^2_{n+1}+F^2_n &F_{n+1}F_{n}+F_{n}F_{n-1}\\ F_{n}F_{n+1}+F_{n-1}F_n &F^2_n+F^2_{n-1}\end{bmatrix} \\ &\equiv \begin{bmatrix}F_{2n+1}&F_{2n}\\F_{2n}&F_{2n-1}\end{bmatrix}=A^{(2n)}.\end{align}$$ So $$\begin{align}F_{2n}&=F_{n+1}F_n+F_{n-1}F_n=(F_{n+1}+F_{n-1})F_n=(F_{n+1}+F_{n-1})(F_{n+1}-F_{n-1})\\&=F^2_{n+1}-F^2_{n-1}.\end{align}$$

edited Mar 14 '18 at 02:09

answered Mar 14 '18 at 00:05

Rócherz

4,241

could you explain the matrix a bit more? does it describe why $F_{2n} =F_{n+1}F_n+F_{n-1}F_n$? – mathkid225 Mar 14 '18 at 00:08
@mathkid225: Yes, sorry. Just finished editing. – Rócherz Mar 14 '18 at 00:11
hmm I don't understand how you made the jump from the 2nd to last matrix to the last matrix – mathkid225 Mar 14 '18 at 00:15
Last matrix is $A^{(2n)}$. – Rócherz Mar 14 '18 at 00:16
just a question, why did you use a matrix? does it have some special properties? sorry, a bit new to matrixes – mathkid225 Mar 14 '18 at 00:20
Matrix $A$ is special because $$\begin{bmatrix}1&1\1&0\end{bmatrix} \begin{bmatrix}F_n\F_{n-1}\end{bmatrix} =\begin{bmatrix}F_n+F_{n-1}\F_n\end{bmatrix} =\begin{bmatrix}F_{n+1}\F_n\end{bmatrix}$$ and computing both $(A^n)^2$ and $A^{(2n)}$ is a fast way (I think) to compare $F_{2n}$ with $F_{n+1}$ and $F_{n-1}$, whose indices differ by $n$ units (minus/plus one). – Rócherz Mar 14 '18 at 00:27
ah, I'm starting to see it thank you! I'll get back if I still don't understand – mathkid225 Mar 14 '18 at 00:28
so if you square $\begin{bmatrix}F_{n+1}\F_n\end{bmatrix}$ you will get $F_{2n}$ correct? – mathkid225 Mar 14 '18 at 02:34
No. Squaring $A^n$ gets you $F_{2n}$. – Rócherz Mar 14 '18 at 03:30

Tal-Botvinnik · Answer 2 · 2018-03-14T00:22:04.430

3

To prove it without induction, you can use the general formula for Fibonacci numbers:

$$F_n=\frac{ \phi_+^n - \phi_-^n}{\sqrt 5}$$

Where $\phi_\pm=\frac{1\pm\sqrt{5}}{2}$.

edited Mar 14 '18 at 00:22

answered Mar 13 '18 at 23:46

Tal-Botvinnik

1,883

I see, so there is no telescope or something we could do? – mathkid225 Mar 13 '18 at 23:47
You mean for the induction? I guess there is a direct way of doing it, using the recurrence relation, taking squares and solving a nasty sum. – Tal-Botvinnik Mar 13 '18 at 23:49
do you have any idea how to complete the induction proof? Still sort of stuck – mathkid225 Mar 13 '18 at 23:55
See the other answer – Tal-Botvinnik Mar 13 '18 at 23:56
1

RHS is missing the $1/\sqrt{5}$. – Rócherz Mar 14 '18 at 00:19
Thanks, corrected – Tal-Botvinnik Mar 14 '18 at 00:21

Prasun Biswas · Answer 3 · 2018-03-14T00:00:55.087

2

A direct proof:

$$F_{k+1}^2-F_{k-1}^2=(F_{k+1}-F_{k-1})(F_{k+1}+F_{k-1})=F_k(F_{k+1}+F_{k-1})=F_kF_{k+1}+F_{k-1}F_k\tag1$$

Summing both sides of (1) over $k$ and using this theorem, we arrive at our required result.

edited Mar 14 '18 at 00:00

answered Mar 13 '18 at 23:55

Prasun Biswas

6,417

score 1 · Answer 4 · answered Mar 14 '18 at 02:03

An alternative method other than induction:

Just by comparison of Fibonacci and Lucas numbers, you can easily find following two identities for all $n\ge 1$: $$L_n=F_{n-1}+F_{n+1}$$ and $$F_{2n}=F_{n}L_{n}$$ The product of these two equations leads to: $$F_{2n}L_n=F_{n}L_{n}\left(F_{n-1}+F_{n+1}\right)$$ Since $L_{n}\ne 0$ when $n\ge 0$, we can cancel $L_{n}$ from both side leading to: $$F_{2n}=F_{n}\left(F_{n-1}+F_{n+1}\right)$$ Now, by substituting $F_{n}=F_{n+1}-F_{n-1}$, we get: $$F_{2n}=\left(F_{n+1}-F_{n-1}\right)\left(F_{n+1}+F_{n-1}\right)=F_{n+1}^2-F_{n-1}^2$$ Done.

How to prove this Fibonacci sequence statement: $F^2_{n+1} - F^2_{n-1} = F_{2n}$

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