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?
$\endgroup$ 24 Answers
$\begingroup$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 $$F_{2n+1}=F_{n+1}^2+F_n^2$$
$\endgroup$ $\begingroup$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 $$ F_{2n} = F_n (F_{n+1} + F_{n-1}), \quad F_{2n+1}=F_n^2+F_{n+1}^2 $$
$\endgroup$ 5 $\begingroup$$$F_{2(n+1)+1}=F_{2n+1}+F_{2n+2}$$
$$=F_{n+1}^2+(F_{n}^2+F_{2n+2})$$
$=F_{n+1}^2+F_{n}^2+F_{n+1}(F_{n+2}+F_n)$ (Using Proving a Fibonacci identity: $F_{2n} = F_n (F_{n+1} + F_{n-1})$,)
$$=F_{n+1}^2+F_{n+1}F_{n+2}+F_n(\underbrace{F_n+F_{n+1}})$$
$$=F_{n+1}^2+F_{n+1}F_{n+2}+F_nF_{n+2}$$
$$=F_{n+1}^2+F_{n+2}(\underbrace{F_{n+1}+F_n})$$
$$=?$$
$\endgroup$ $\begingroup$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. $\square$
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