How can I prove that: $\frac{(n+1)(n+2)(n+3)...(3n)}{(1*2)(4*5)(7*8)...(3n-2)*(3n-1)} = 3^n$

Question

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.

Answer 1

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

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

Answer 2

$$\frac{(n+1)(n+2)(n+3)...(3n)}{(1\cdot2)(4\cdot5)(7\cdot8)...(3n-2)\cdot(3n-1)}=$$ $$=\frac{3^n\cdot n!(n+1)(n+2)(n+3)...(3n)}{(1\cdot2)3(4\cdot5)6(7\cdot8)...(3n-2)\cdot(3n-1)3n}=3^n$$