<s>[INST] Consider sequence of numbers $a_r,\;r\geq0\;$ with $a_0=1\;$ and $a_{r+1}^2=1+a_r\cdot a_{r+2}.\;$ Then which of the following is/are true? Let $\alpha, \beta$ are roots of equation $x^2-a_1 x+1=0$ and consider sequence of numbers $a_r,\;r\geq0\;$ with $a_0=1\;$ and $a_{r+1}^2=1+a_r\cdot a_{r+2}.\;$ Then which of the following is/are true?
(A) $a_r+a_{r+2}=a_1 \cdot a_{r+1}$
(B) $a_r+a_{r+2}=2\cdot a_{r+1}$
(C) $a_n=\dfrac{\alpha^{n+1}-\beta^{n+1}}{\alpha-\beta}$
(D) $a_n=\dfrac{\alpha^{n+1}+\beta^{n+1}}{\alpha+\beta}$
My Approach:
Since it is given that  $a_{r+1}^2=1+a_r\cdot a_{r+2}.\; \implies  a_{r+1}^2-1=a_r\cdot a_{r+2}.\;$
Now using option (A)   $\;a_r+a_{r+2}=a_{r}+\dfrac{a_{r+1}^2-1}{a_r}\implies a_r+a_{r+2}=\dfrac{a_{r}^2-1+a_{r+1}^2}{a_r}\implies a_r+a_{r+2}=\dfrac{(a_{r-1}+a_{r+1})\cdot a_{r+1}}{a_r}.$
Now $a_{r-1}+a_{r+1}=a_{r-1}+\dfrac{a_{r}^2-1}{a_{r-1}}\implies a_{r-1}+a_{r+1}=\dfrac{a_{r-1}^2-1+a_{r}^2}{a_{r-1}}\implies a_{r-1}+a_{r+1}=\dfrac{(a_{r-2}+a_{r})\cdot a_{r}}{a_{r-1}}$
When i continue this process i obtain $a_{r}+a_{r+2}=\dfrac{(a_0+a_2)a_2}{a_1}$
But answer given is (A), How do i obtain answer as option (A) which depend upon $\;a_{r+1}?\;$
Also how to obtain option (C) and (D)?
 [/INST] Hint: The following are rather hints, as we don't prove claims for each $n$ resp. $r$. The solutions of the quadratic equation $x^2-a_1 x+1=0$ are
\begin{align*}
\alpha=\frac{1}{2}\left(a_1+\sqrt{a_1^2-4}\right)\qquad\qquad
\beta=\frac{1}{2}\left(a_1-\sqrt{a_1^2-4}\right)\tag{1}
\end{align*}
Point (C):
With $n=1$ we obtain
\begin{align*}
a_1=\frac{\alpha^2-\beta^2}{\alpha-\beta}=\alpha+\beta
\end{align*}
which coincides with (1) indicating (C) is correct.
Point (D):
With $n=1$ we obtain
\begin{align*}
a_1&=\frac{\alpha^2+\beta^2}{\alpha+\beta}\\
&=\frac{\frac{1}{4}\left(a_1^2+2a_1\sqrt{a_1^2-4}+a_1^2-4\right)+\frac{1}{4}\left(a_1^2-2a_1\sqrt{a_1^2-4}+a_1^2-4\right)}{a_1}\\
&=\frac{a_1^2-2}{a_1}
\end{align*}
It follows $a_1^2=a_1^2-2$ which is absurd and (D) is not correct.
Point (A):
With $r=0$ we obtain
\begin{align*}
a_0+a_2=a_1^2
\end{align*}
and since $a_0=1$ we have
\begin{align*}
a_1^2=1+a_2
\end{align*}
The recurrence relation $a_{r+1}^2=1+a_r\cdot a_{r+2}$ gives also for $r=0$
\begin{align*}
a_1^2=1+a_0a_2=1+a_2
\end{align*}
which indicates (A) is correct.


Addendum (2022-06-10): Proof of (A)
Since OP has already shown the validity of (C) which is stated in a comment of this post
\begin{align*}
\text{(C)}\qquad a_r=\frac{\alpha^{r+1}-\beta^{r+1}}{\alpha-\beta}\qquad\qquad r\geq 0\tag{1}
\end{align*}
we will conveniently use it to prove (A). We also recall from the quadratic equation
\begin{align*}
x^2-a_1x+1&=0\\
\alpha+\beta&=a_1\tag{2}\\
\alpha\beta&=1\tag{3}
\end{align*}

We obtain for $r\geq 0$:
\begin{align*}
\color{blue}{a_1a_{r+1}}&=\left(\alpha+\beta\right)\frac{\alpha^{r+2}-\beta^{r+2}}{\alpha-\beta}\tag{$\rightarrow (1),(2)$}\\
&=\frac{\alpha^{r+3}+\beta\alpha^{r+2}-\alpha\beta^{r+2}-\beta^{r+3}}{\alpha-\beta}\\
&=\frac{\left(\alpha\beta\right)\alpha^{r+1}-\left(\alpha\beta\right)\beta^{r+1}}{\alpha-\beta}
+\frac{\alpha^{r+3}-\beta^{r+3}}{\alpha-\beta}\tag{$\rightarrow (3)$}\\
&\,\,\color{blue}{=a_r+a_{r+2}}
\end{align*}
and the claim (A) follows.

</s>