Identity of Fibonacci sequence

Question

Let $F_n$ be a Fibonacci sequence with initial terms $F_0=0, F_1=1$ and $F_{n+1}=F_n+F_{n-1}$ for $n\geqslant 1$.

Prove that $F_n^2+F_{n+1}^2=F_{2n+1}$ for $n\geqslant 0$ (with mathematical induction).

My efforts: For $n=0$ it is true.

Suppose that our statement holds for $0\leqslant k \leqslant n$ i.e. $F_k^2+F_{k+1}^2=F_{2k+1}$

Let's try to prove it for $k=n+1$. $$F_{2n+3}=F_{2n+1}+F_{2n+2}=F_{n+1}^2+(F_{n}^2+F_{2n+2})= ?$$

Here I'm stuck and I have applied different methods but none of them brings a positive result.

Can anyone help to complete this?

Answer 1

We can prove more general identity, i.e. $F_{n+m+1}=F_{m+1}F_{n+1}+F_{m}F_{n}$.

Let's prove above by mathematical induction on $n$.

For $n=0$ we have $F_{m+1}=F_{m+1}$

Suppose it is true for all $0\leqslant n \leqslant k$.

Let's prove it for $n=k+1$ then $$F_{k+m+2}=F_{k+m}+F_{k+m+1}=(F_{m+1}F_{k}+F_{m}F_{k-1})+(F_{m+1}F_{k+1}+F_{m}F_{k})=F_{m+1}(F_k+F_{k+1})+F_m(F_{k-1}+F_k)=F_{m+1}F_{k+2}+F_mF_{k+1}$$

The first two parentheses were derived from induction assumption for $n=k-1$ or $n=k$.

Thus, we have proved our initial statement. If we put $n=m$ we get $$F_{2n+1}=F_{n+1}^2+F_n^2$$

Answer 2

This follows from the matrix formulation, which is well worth knowing and easily proved: $$ \begin{pmatrix}1&1\\1&0\end{pmatrix}^n= \begin{pmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix} $$ Just compare $$ \begin{pmatrix}1&1\\1&0\end{pmatrix}^{2n}= \begin{pmatrix}F_{2n+1}&*\\*&*\end{pmatrix} $$ with $$ \begin{pmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix}^2= \begin{pmatrix}\cdots&*\\*&*\end{pmatrix} $$ Alternatively (and equivalently), you can use induction to prove simultaneously that $$ F_{2n} = F_n (F_{n+1} + F_{n-1}), \quad F_{2n+1}=F_n^2+F_{n+1}^2 $$

Answer 3

$$F_{2(n+1)+1}=F_{2n+1}+F_{2n+2}$$

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

$=F_{n+1}^2+F_{n}^2+F_{n+1}(F_{n+2}+F_n)$ (Using Proving a Fibonacci identity: $F_{2n} = F_n (F_{n+1} + F_{n-1})$,)

$$=F_{n+1}^2+F_{n+1}F_{n+2}+F_n(\underbrace{F_n+F_{n+1}})$$

$$=F_{n+1}^2+F_{n+1}F_{n+2}+F_nF_{n+2}$$

$$=F_{n+1}^2+F_{n+2}(\underbrace{F_{n+1}+F_n})$$

$$=?$$

Answer 4

We can actually prove two different identities (1) and (2), of which you were asked to prove (1): $$F_{n+1}^2+F_{n+2}^2=F_{2n+3} \quad \text{(1)}$$ $$F_{n+2}^2-F_{n}^2=F_{2n+2} \quad \text{(2)}$$ Note that the Fibonacci numbers are numbered starting from $F_1$.

We can proceed by induction. Suppose that we have the two equations $$F_n^2+F_{n+1}^2=F_{2n+1} \quad \text{(3)}$$ $$F_{n+1}^2-F_{n-1}^2=F_{2n} \quad \text{(4)}$$ The base cases are trivial.

First Step: Prove (2).

Now, adding (3) and (4), we obtain : \begin{align} 2F_{n+1}^2+F_n^2-F_{n-1}^2 &= F_{2n+1}+F_{2n}\\\\ (F_{n+1}-F_n)^2+F_{n+1}^2 + 2F_{n+1}F_n - F_{n-1}^2&= F_{2n+2}\\\\ (F_{n-1})^2+F_{n+1}^2 + 2F_{n+1}F_n - F_{n-1}^2&= F_{2n+2}\\\\ (F_{n+1}+F_{n})^2-F_{n}^2&= F_{2n+2}\\\\ F_{n+2}^2-F_{n}^2&= F_{2n+2} \quad \text{(2)} \end{align} As desired. $\square$

Second Step: Prove (1). Now that we have (2), we are home free!

Adding (2) and (3), we obtain :

\begin{align} F_{n+2}^2-F_n^2+F_{n}^2+F_{n+1}^2 &= F_{2n+2}+F_{2n+1}\\\\ F_{n+2}^2+F_{n+1}^2 &= F_{2n+3} \quad \text{(1)} \end{align} And we are done. $\square$