From $a_{n+1}= \frac{3a_n}{(2n+2)(2n+3)}$ to $a_n$, Case 2 
Find and prove by induction an explicit formula for $a_n$ if $a_1=1$ and, for $n \geq 1$,
$$P_n:  a_{n+1}= \frac{3a_n}{(2n+2)(2n+3)}$$

Checking the pattern:
$$a_1=1 $$
$$a_2= \frac{3}{4 \cdot 5}$$
$$a_3= \frac{3^2} { 4 \cdot 5 \cdot 6 \cdot 7}$$
$$a_4= \frac{3^3} { 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 }$$
$$a_n = \frac{3^{n-1}}{ \frac {(2n+1)!}{3!} }$$
$$a_n = \frac {3! \cdot 3^{n-1}} {(2n+1)!}$$
Proof by induction:
$$a_1 = \frac {3! \cdot 3^{0}} {3!} =1$$
$$a_{n+1} = \frac {3! \cdot 3^{n}} {(2n+3)!}$$
Very grateful for the feedback given before. I am new to this. I am a bit stock at the end, what is the most efficient way to reach back to $a_n$?

$$a_{n+1} = \frac{n!.3^n}{(2n+3)!} = \frac{n!.3.3^{n-1}}{(2n+3)(2n+2)(2n+1)!}=\frac{3!3^{n-1}}{(2n+1)!} \frac{3}{(2n+2)(2n+3)} = a_n \frac{3}{(2n+2)(2n+3)} = \frac{3a_n}{(2n+2)(2n+3)}$$
