Prove that $a_{n+1}=\frac{a_n^2+4}{a_{n-1}}$ recursion formula is equivalent to $\frac{a_{n+1}+a_{n-1}}{a_n}=6$ if $a_1=1, a_2=5$ $\forall n \gt1$

Question

Prove that $a_{n+1}=\frac{a_n^2+4}{a_{n-1}}$ recursion formula is equivalent to $\frac{a_{n+1}+a_{n-1}}{a_n}=6 \hspace{0,5cm}$

if $a_1=1, a_2=5$ $\forall n \gt1$

I spent a couple of hours with the task, which can be found on the below link, when I noticed the equivalence between the two recursion formulas, unfortunately I could not prove it.

Finding the general formula of $a_{n+1}=\frac{a_n^2+4}{a_{n-1}}$ with $a_1=1$ and $a_2=5$

Could someone help with the proof, thanks in advance.

Answer 1

To give an induction proof, suppose that

$$a_{n+1}={a_n^2+4\over a_{n-1}}=6a_n-a_{n-1}$$

which holds for the base case, $n=1$ when $a_0=1$ and $a_1=5$, since $(5^2+4)/1=29=6\cdot5-1$. We want to prove that $(a_{n+1}^2+4)/a_n=6a_{n+1}-a_n$, in which case we can say both are equal to $a_{n+2}$.

From $a_{n+1}=6a_n-a_{n-1}$ we have

$$0=a_{n+1}(a_{n+1}-6a_n+a_{n-1})=a_{n+1}^2-6a_na_{n+1}+a_{n+1}a_{n-1}$$

From $a_{n+1}a_{n-1}=a_n^2+4$, this becomes

$$0=a_{n+1}^2-6a_na_{n+1}+a_n^2+4=a_{n+1}^2+4-a_n(6a_{n+1}-a_n)$$

and thus $(a_{n+1}^2+4)/a_n=6a_{n+1}-a_n$, as desired.

Answer 2

Starting from the second recusion formula $a_{n+1}=6a_n-a_{n-1}$, we get for all $n\ge2$ $$ \left[\begin{array}{cc}a_{n+2}&a_{n+1}\\a_{n+1}&a_{n}\end{array}\right]=\left[\begin{array}{cc}6&-1\\1&0\end{array}\right]\left[\begin{array}{cc}a_{n+1}&a_n\\a_n&a_{n-1}\end{array}\right]. $$ Define $\displaystyle M_n=\left[\begin{array}{cc}a_{n+1}&a_n\\a_n&a_{n-1}\end{array}\right]$ and $\displaystyle A=\left[\begin{array}{cc}6&-1\\1&0\end{array}\right]$. Observe that we have $$ \det(M_{n+1})=\det(A)\det(M_n)=\det(M_n)=\cdots=\det(M_2)=4. $$ Since $\det(M_n)=a_{n+1}a_{n-1}-a_n^2=4$ is equivalent to the first recursion formula, we can see that the two formulae give the same sequence because the first two terms determine the whole sequence.

Note: It can be noted that this relationship can be generalized to $a_{n+1}= ta_n-a_{n-1}$ and $a_{n+1}a_{n-1}-a_n^2=rst-r^2-s^2$ with the common first two terms $a_2=s, a_1=r$ and arbitrary $t$.