<s>[INST] How can I prove that: $\frac{(n+1)(n+2)(n+3)...(3n)}{(1*2)(4*5)(7*8)...(3n-2)*(3n-1)} = 3^n$ How can I prove that:
$$\frac{(n+1)(n+2)(n+3)...(3n)}{(1 * 2)(4*5)(7*8)...(3n-2)*(3n-1)} = 3^n$$  
Can you help me and explain me how can I prove it? I thought to prove it by induction, but I don't have idea how to do it in fact.
 [/INST] Note that


*

*$(n+1)(n+2)(n+3)…(3n)=\frac{(3n)!}{n!}$

*$(1\cdot2)(4\cdot5)(7\cdot8)…(3n-2)\cdot(3n-1)=\frac{(3n-1)!}{3\cdot6\cdot9\cdot...\cdot(3n-3)}=\frac{(3n-1)!}{3^{n-1}(n-1)!}$


thus
$$\frac{(n+1)(n+2)(n+3)…(3n)}{(1\cdot2)(4\cdot5)(7\cdot8)…(3n-2)\cdot(3n-1)} = \frac{(3n)!}{n!}\frac{3^{n-1}(n-1)!}{(3n-1)!}=\\=3^{n-1}\frac{(3n)!}{(3n-1)!}\frac{(n-1)!}{n!}=3^{n-1}\cdot\frac{3n}n=3^n$$
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