A recurrence relation for Fibonacci squared

Question

The sequence $S_n$ is defined as $$S_1=S_2 =1$$ and for $n\ge 2$, $$S_{n+1}=2(A_n + G_n)$$ where $A_n = \frac {S_n+S_{n-1}}{2}$ is the arithmetic mean and $G_n= \sqrt { S_nS_{n-1} }$ is the geometric mean.

Claim: $$S_n = (F_n)^2$$ where $F_n$ stands for Fibonacci's numbers.

My solution:

$$S_{n+1}=2(A_n +G_n)= 2(\frac {S_n+S_{n-1}}{2}+\sqrt { S_nS_{n-1} })$$

$$=S_n+S_{n-1}+2\sqrt { S_nS_{n-1} }$$

$$= ( \sqrt { S_n} + \sqrt S_{n-1} )^2$$

$$\sqrt { S_{n+1}}=\sqrt { S_n} + \sqrt S_{n-1} \implies $$

$$\sqrt { S_n} =F_n \implies S_n = (F_n)^2$$

Answer 1

$$S_{n+1}=S_n+2\sqrt{S_nS_{n-1}}+S_{n-1}=(\sqrt{S_n}+\sqrt{S_{n-1}})^2$$ So if we let $F_n=\sqrt{S_n}$, then $F_1=F_2=1$ and $$F_{n+1}=F_n+F_{n-1}$$ Can you take it from here?

Answer 2

You have found "A recurrence relation for Fibonacci squared" (your title) by a certain kind of approach.

Here is another approach using a third order recurrence relationship for sequence $(F_n^2)$ :

$$F_n^2=2F_{n-1}^2+2F_{n-2}^2-F_{n-3}^2\tag{1}$$

with, of course, initial values

$$F_1^2=1, \ F_2^2=1, \ F_3^2=4.$$

Proof of (1) :

$$\begin{cases}F_{n}=F_{n-1}+F_{n-2} &\implies&F_{n}^2=(F_{n-1}+F_{n-2})^2\\ F_{n-3}=F_{n-1}-F_{n-2} &\implies&F_{n-3}^2=(F_{n-1}-F_{n-2})^2\end{cases}$$

Adding these relationships gives (1).

(1) can be used in order to find an explicit formula for $F_n^2$.

Indeed, the characteristic equation associated with (1) is

$$r^3-2r^2-2r+1=(r+1)(r^2-3r+1)$$

Its roots are

$$r_1=(-1), \ r_2=\tfrac12(3+\sqrt{5}), \ r_3=\tfrac12(3-\sqrt{5})\tag{2}$$

Taking into account initial conditions, the explicit solution is :

$$F_n^2=-\dfrac25 r_1^n + \dfrac15 r_2^n + \dfrac15 r_3^n \tag{3}$$

Remarks :

a) (3) could have been as well obtained by squaring Binet relationship:

$$F_n=\dfrac{1}{\sqrt{5}}\left(\left(\tfrac{1+\sqrt{5}}{2}\right)^n-\left(\tfrac{1-\sqrt{5}}{2}\right)^n\right).$$

b) An interesting generalization of (1) to higher powers of $F_n$ can be found in : http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/Fibonomials.html#squares