Solve $a_{n+2}=a_{n} \times \frac{(n+1)(n+4)}{(n+2)(n+3)}$

Question

How to solve this recurrence relation ? $$a_{n+2}=a_{n} \times \frac{(n+1)(n+4)}{(n+2)(n+3)}$$

I am new to recurrence solving.

I found somewhere that the solution of $$a_{n}=f(n)a_{n-1}$$ is $$a_{n}=a_{0}\prod_{r=0}^{n-1}f(r)$$ Is this method is applicable here ?

Answer 1

Similar to the result you found, if $a_{n+2}=\frac{f(n+2)}{f(n)}a_n$ then, for $j\in\{0,\,1\}$, you may prove by induction on $n$ that $a_{2n+j}=\frac{f(2n+j)}{f(j)}a_j$. In your case, take $f(n)=\frac{n+2}{n+1}$. Notice that we need $a_k$ for one even $k$ and one odd $k$, e.g. $a_0$ and $a_1$, to specify the sequence.

Answer 2

What is amazing in your problem is that defining $$a_n= \frac{n+2 } {n+1 }\,b_n$$ you just end with $$b_{n+2}=b_n$$ which is quite pleasant (I hope).