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.
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Jaideep Khare
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You can do it by induction. Ask yourself, how does the left-hand side change when you go from $n$ to $n+1$? How does the numerator change? How does the denominator change? – saulspatz Feb 04 '18 at 21:41
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1If you "WantToKnowMath", then you gotta start doing math, and stop always asking others to do your math assignments for you once in a while! – amWhy Feb 04 '18 at 22:52
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$$\frac{(n+1)(n+2)(n+3)...(3n)}{(1\cdot2)(4\cdot5)(7\cdot8)...(3n-2)\cdot(3n-1)}=$$ $$=\frac{3^n\cdot n!(n+1)(n+2)(n+3)...(3n)}{(1\cdot2)3(4\cdot5)6(7\cdot8)...(3n-2)\cdot(3n-1)3n}=3^n$$
Michael Rozenberg
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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
$$\frac{(n+1)(n+2)(n+3)…(3n)}{(1\cdot2)(4\cdot5)(7\cdot8)…(3n-2)\cdot(3n-1)} = \frac{(3n)!}{n!}\frac{3^{n-1}(n-1)!}{(3n-1)!}=\\=3^{n-1}\frac{(3n)!}{(3n-1)!}\frac{(n-1)!}{n!}=3^{n-1}\cdot\frac{3n}n=3^n$$
user
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