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If I sum the terms of a binomial expansion, which would be the least common multiple of all the denominators? Say

$\displaystyle \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \ldots + \binom{n}{n}$

$1 \displaystyle + \frac{n!}{(n - 1)!} + \frac{n!}{2!(n - 2)!} + \ldots + 1$

$n$ may be odd or even (if $n$ is even, when $k = n/2$ we have a $(n - n/2)^2$ denominator).

I modify the original post in order to better understand my question. "What is $\operatorname{lcm} \{ k!(n - k)! \}$ for $k = 0, \ldots, n$"?

Do you have any hints?

Thank you anyway!

Michael Hardy
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BowPark
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    $$\sum_{r=0}^n\binom nr= (1+1)^n=2^n$$ and also http://math.stackexchange.com/questions/11601/proof-that-a-combination-is-an-integer – lab bhattacharjee Feb 28 '14 at 18:40
  • Thank you. But maybe your hint was referred to the result of the sum? (Because I don't need it, but just the l.c.m. and so the addends in the numerator). – BowPark Feb 28 '14 at 18:50
  • Ok, it is right, but see in the answer below: my question is "what is $lcm { k!(n - k)! }$ for $k = 1, \cdots, n$"? – BowPark Feb 28 '14 at 19:11
  • Actually it is important that $k$ start from zero or one. If it starts from zero the answer is simply $n!$, but if it starts from 1, then $n!$ is not the answer any more, for example if $n$ is prime and $k$ starts from 1 then the answer is $(n-1)!$, so make your mind "zero or one?"!! – Woria Feb 28 '14 at 19:22
  • Ok, say $k$ starts from $0$. But are you sure the lcm in that case is $n!$? There are terms like $(n - n/2)^2$ (if $n$ is even) – BowPark Feb 28 '14 at 19:25
  • Yes, it's $n!$ since $k!(n-k)!$ divides $n!$ which is because $\binom nk$ is an integer. – user2345215 Feb 28 '14 at 19:33
  • If $k$ starts from $0$, then we have $n!$ in that set, and the lcm should be a multiple of $n!$. Other numbers in that set are of the form $k!(n-k)!$ and $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ is an integer, i.e $k!(n-k)!|n!$, thus the lcm is $n!$. – Woria Feb 28 '14 at 19:34

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In every case the fraction reduced to lowest terms has denominator $1$, and $\operatorname{lcm}(1,1,1,\ldots,1)=1$.

Michael Hardy
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