Prove by mathematical induction that $n(n+1)(n+2)...(n+r-1)$ is divisible by $r!$. I tried by proving it true for $r=1$ and then assuming it true for $r=k$ and proving it true for $r=k+1$ But I could not prove that. Do I need to do the same thing for n?
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@Arthur my mistake, thought it was $n(n-1)\dots (n-r+1)$ – Mark Schultz-Wu Dec 24 '16 at 07:47
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http://math.stackexchange.com/questions/12065/the-product-of-n-consecutive-integers-is-divisible-by-n-factorial http://math.stackexchange.com/questions/12067/the-product-of-n-consecutive-integers-is-divisible-by-n-without-using-the-prop – lab bhattacharjee Dec 24 '16 at 08:51
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To save a lot of typing, let me define $$f(n,r)=n(n+1)(n+2)\cdots(n+r-1).$$ You can verify the identity $$f(n,r+1)=(r+1)f(n,r)+f(n-1,r+1)$$ which you can then use to prove the statement $$f(n,r)\text{ is divisible by }r!$$ by induction on the quantity $n+r.$
Alternatively, if you're allowed to recast the problem in terms of binomial coefficients, you can prove that $$\frac{f(n,r)}{r!}=\binom{n+r-1}r$$ is an integer by using Pascal's identity $$\binom x{r+1}=\binom{x-1}r+\binom{x-1}{r+1}$$ to prove by induction on $m$ that $\binom mr$ is an integer for all integers $m,r\ge0.$ (Of course this is just the same proof in different notation.)
bof
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