1. ## Fibonacci Sequence property

Using normal properties of Fibonacci Numbers Prove that:
(F(n+1)^2)+(F(n)^2)= F(2n+1).
We see that F(3)=2 and F(4)=3 and so F(7)=13 so it is true
Not sure where to start here and any help would be appreciated.

2. Originally Posted by MatsSundin
Using normal properties of Fibonacci Numbers Prove that:
(F(n+1)^2)+(F(n)^2)= F(2n+1).
We see that F(3)=2 and F(4)=3 and so F(7)=13 so it is true
Not sure where to start here and any help would be appreciated.
i suppose you are doing induction.

so, you must now assume P(n) is true and try to show that P(n + 1) must be true.

of course, here, $P(n)$ is " $F_{n + 1}^2 + F_n^2 = F_{2n + 1}$ for all integers $n \ge 3$"

3. I proved this identity back in school using Binet's formula(without induction) and some identities.

Do you have to use induction?. That is a good way to go, but I like this method I stumbled onto a while back while studying Fibonacci's for a class called Seminars in Mathematics. I doubt if it's original, but it works.

I will use $x=\frac{1+\sqrt{5}}{2}, \;\ y=\frac{1-\sqrt{5}}{2}$

Take note that $xy=-1$ and $x^{2}-\sqrt{5}x+1=0$ and $y^{2}+\sqrt{5}y+1=0$

We will use these later.

From Binet's formula: $\frac{x^{n}-y^{n}}{\sqrt{5}}$

We can sub it in our formula to prove:

$\frac{1}{5}\left[(x^{n}-y^{n})^{2}+(x^{n+1}-y^{n+1})^{2}-\sqrt{5}(x^{2n+1}-y^{2n+1})\right]$

Expand:

$\frac{1}{5}\left[x^{2n}-2x^{n}y^{n}+y^{2n}+x^{2n+2}-2x^{n+1}y^{n+1}+y^{2n+2}-\sqrt{5}x^{2n+1}+\sqrt{5}y^{2n+1}\right]$

Group and factor:

$\frac{1}{5}\left[x^{2n}(\underbrace{x^{2}-\sqrt{5}x+1}_{\text{equals 0}})+y^{2n}(\underbrace{y^{2}+\sqrt{5}y+1}_{\text{ equals 0}})-2(\underbrace{xy}_{\text{-1}})^{n}(1+xy)\right]$

Take note that everything inside the brackets is 0, due to the identities previously mentioned.

We get:

$\frac{1}{5}\left[x^{2n}(0)+y^{2n}(0)-2(-1)(0)\right]=0$

$\therefore$, $f^{2}_{n}+f^{2}_{n+1}=f_{2n+1}$

You can show various identities using these tools. They can be handy to know.

4. No induction is not needed, it can be proven any way possible. Your solution is quite intuitive and impressive thank you!