<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] Proof for (A) follows. I've pretty sure induction is the best method; you're already dealing with recurrences, why not add in one more?
I'm going to designate $a_1$ as just $a$ for the sake of simplicity, because it'll show up a lot more than the other $a_r$.
As you already showed, $a_{r+2} = (a_{r+1}^2 - 1)/a_r$. For $r=0$, this is just $a_2 = (a^2-1)/1 = a^2 - 1$.
Since we have an expression in just $a$ for $a_2$, we can see $a_3 = ((a^2-1)^2-1)/a = a^3 - 2a$. We could continue this*, but it's not necessary.
First, we can see quickly that (A) is true for $r=0,1$:
$$a_2 + a_0 = a^2-1+1 = a^2 = a_1 \cdot a_1 \\
a_3 + a_1 = a^3-2a+a = a(a^2-1) = a_1 \cdot a_2$$
Let's do some induction. We want to show that:
$$a_r + a_{r+2} = a \cdot a_{r+1} \implies a_{r+1} + a_{r+3} = a \cdot a_{r+2}$$
Or, having shown $P(0), P(1)$ to be true, we must show that $P(r) \implies P(r+1)$.
I hate big subscripts**, so I'm setting $x = a_r, \ y= a_{r+1}, \ z= a_{r+2}, \ w = a_{r+3}$. This means we want to show:
$$x+z=ay \implies y+w=az$$
Let's begin by ridding ourselves of the $w$. We know $w = (z^2 - 1)/y$, and we can rewrite the RHS of our implication as $y^2+z^2-1 = ayz$. But we also know that $y^2-1=xz$. Making that substitution gives us a RHS of $xz+z^2=ayz$
Now, none of these values can be $0$, as $a_r = 0 \implies a_{r+2} = (a_r^2-1)/0$, and division by zero is a nicely instant contradiction. Since $z \ne 0$, we can divide by it to get $x+z=ay$ on the RHS, which is also the LHS of our implication. Since the LHS and RHS are the same, the implication must be true, and thus $P(r) \implies P(r+1)$.
Because $P(0)$ and $P(1)$ are true, and $P(r) \implies P(r+1)$ is true, $P(r)$ is true for all $r \ge 0$, and statement (A) is proven. $\square$
(Note the importance of not using the LHS as a substitution in the RHS, which would cause logical loops.)

*If we had continued with the resubstitutions, we actually get a very nice pattern:
$$
\begin{array}{l l l l l}
a_0 = & 1 \\
a_1 = & a \\
a_2 = & a^2 & -1 \\
a_3 = & a^3 & -2a \\
a_4 = & a^4 & -3a^2 & +1 \\
a_5 = & a^5 & -4a^3 & +3a \\
a_6 = & a^6 & -5a^4 & +6a^2 & -1 \\
a_7 = & a^7 & -6a^5 & +10a^3 & -4a
\end{array}
$$
The coefficients, if you read going southeast, are the rows of Pascal's Triangle. So column $2$ is the integers, column $3$ is the triangular numbers, etc. It sure looks like each line is:
$$a_r = \binom{r}{0}a^r - \binom{r-1}{1}a^{r-2} + \binom{r-2}{2} a^{r-4} - \binom{r-3}{3}a^{r-6} + \cdots = \sum_{i=0}^{\lfloor r/2 \rfloor} (-1)^i \binom{r-i}{i}a^{r-2i}$$
...but I'm not going to try proving it.
**$\tiny\text{(And I cannot lie)}$
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