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: Plug it back to formula $P_n:  a_{n+1}= \frac{3a_n}{(2n+2)(2n+3)}$. By induction hypothesis, $a_n = \frac {3! \cdot 3^{n-1}} {(2n+1)!}$, so 
$$a_{n+1}= \frac{3 \cdot 3! \cdot 3^{n-1}}{(2n+2)(2n+3)\cdot (2n+1)!} = \frac{3! \cdot 3^{n}}{(2n+3)!}$$
which is what you need to complete the induction.
A: $a_{n+1} = \frac{3a_n}{(2n+2)(2n+3)} = \frac{3\cdot 3!\cdot 3^{n-1}}{(2n+2)(2n+3)(2(n-1)+3)!} = \frac{3!3^n}{(2n+3)!}$.
A: $$a_{n+1} = \frac{n!.3^n}{(2n+3)!} = \frac{n!.3.3^{n-1}}{(2n+3)(2n+2)(2n+1)!}=\frac{3!3^{n-1}}{(2n+1)!} \frac{3}{(2n+2)(2n+3)} = a_n \frac{3}{(2n+2)(2n+3)} = \frac{3a_n}{(2n+2)(2n+3)}$$
A: You assume the expression to be true for $n=m$ and then prove it is true for $n=m+1$.
Expression:
\begin{gather}
a_n = \frac {3! \cdot 3^{n-1}} {(2n+1)!}\\
\\
\text{For } n = m \text{ assume true}\\
\\
a_{m+1}= \frac{3a_m}{(2m+2)(2m+3)} \\
a_{m+1}= \frac{3}{(2m+2)(2m+3)}\cdot \frac {3! \cdot 3^{m-1}} {(2m+1)!}\\
a_{m+1}= \frac {3! \cdot 3^{m}} {(2m+3)!} = \frac {3! \cdot 3^{(m+1)-1}} {(2(m+1)+1)!}
\end{gather}
And proved by the principle of MI. 
Edit: To elaborate a bit on this, what you've done is that you've shown for a base case, that the expression is true. Then you've picked a arbitrary index (in this case $m$) and shown it's true for $m+1$. This way you've proved that if $1$ is true, then so is $2$ and if $2$ is true so is $3$ and so on, because $m$ is arbitrary. 
