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?

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    $\begingroup$ math.stackexchange.com/questions/295173/… $\endgroup$ – lab bhattacharjee Nov 28 '17 at 11:14
  • $\begingroup$ @labbhattacharjee, thanks but your link has matrix proof which is given below and proof suing binet's formula. As I said above i am interested in induction proof $\endgroup$ – ZFR Nov 28 '17 at 11:18

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 $$ F_{2n} = F_n (F_{n+1} + F_{n-1}), \quad F_{2n+1}=F_n^2+F_{n+1}^2 $$

  • $\begingroup$ Adapted from math.stackexchange.com/a/1505284/589 $\endgroup$ – lhf Nov 28 '17 at 11:14
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    $\begingroup$ Nice proof but I would like to see the induction proof $\endgroup$ – ZFR Nov 28 '17 at 11:15
  • $\begingroup$ How to use induction in order to prove the second identity? Could you write the induction proof? $\endgroup$ – ZFR Nov 28 '17 at 11:34
  • $\begingroup$ @RFZ, proceed as in lab bhattacharjee's answer. $\endgroup$ – lhf Nov 28 '17 at 12:37
  • $\begingroup$ I have added an answer. Please take a look $\endgroup$ – ZFR Nov 28 '17 at 15:21

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




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






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