USA TST 2001 #2
by Wolstenholme, Oct 14, 2014, 11:45 PM
Express ![\[ \sum_{k=0}^n (-1)^k (n-k)!(n+k)! \]](//latex.artofproblemsolving.com/7/b/8/7b85fdabf6aae54c00ba6a5f351729a4e15887fb.png)
in closed form.
Solution:
Let
be the Beta function (or an Eulerian integral of the first kind).
Lemma:
Proof:
Note that
as desired.
Now, returning to the original problem, note that
![\[ (2n + 1)!\sum_{k = 0}^{n}(-1)^kB(n + k, n - k) = \]](//latex.artofproblemsolving.com/9/e/7/9e7a3f58ef4cdb7596002c314f8705b8c9e41fed.png)
![\[ (2n + 1)!\sum_{k = 0}^{n}(-1)^k\int_{0}^{1}t^{n + k}(1 - t)^{n - k}dt = \]](//latex.artofproblemsolving.com/3/2/f/32fe91a90eb86a20d8613ee61519785971846606.png)
![\[ (2n + 1)!\sum_{k = 0}^{n}(-1)^k\int_{0}^{1}\sum_{j = 0}^{n - k}\binom{n - k}{j}(-1)^jt^{n + k +j} dt = \]](//latex.artofproblemsolving.com/8/8/a/88a5863fd484430a9520588b75880c001d23cd39.png)
![\[ (2n + 1)!\sum_{k = 0}^{n}(-1)^k\sum_{j = 0}^{n - k}\frac{(-1)^j}{n + k + j + 1}\binom{n - k}{j} = \]](//latex.artofproblemsolving.com/0/e/e/0eed69d958c7893712cb46d0571f2bff3329d5bc.png)
![\[ (2n + 1)!\sum_{m = 0}^{n}\frac{(-1)^m}{n + m + 1}\sum_{j + k = m}\binom{n - k}{j} = \]](//latex.artofproblemsolving.com/e/2/7/e27c1860b972087c4a562a76bbcd91939b4aa09e.png)
![\[ (2n + 1)!\sum_{m = 0}^{n}\frac{(-1)^m}{n + m + 1}\binom{n + 1}{m} = \]](//latex.artofproblemsolving.com/4/d/e/4de4e4f37cb184ce86bb36a78c693c848587ec97.png)
![\[ (2n + 1)!\left(B(n, n + 1) + \frac{(-1)^n}{2(n + 1)}\right) = \]](//latex.artofproblemsolving.com/0/3/e/03e5c0b7ac7d3567a432bb1733f6b2b64df976bb.png)
![\[ (2n + 1)!\left(\frac{n!(n + 1)!}{(2n + 2)!} + \frac{(-1)^n}{2n + 2}\right) \]](//latex.artofproblemsolving.com/4/7/4/47439f1dc43b462b3f1410157ee450070b20323f.png)
And so we are done. Essentially this problem reduced to two applications of the binomial theorem on the integrand of the Beta function.
![\[ \sum_{k=0}^n (-1)^k (n-k)!(n+k)! \]](http://latex.artofproblemsolving.com/7/b/8/7b85fdabf6aae54c00ba6a5f351729a4e15887fb.png)
in closed form.Solution:
Let
be the Beta function (or an Eulerian integral of the first kind).Lemma:
Proof:
Note that
as desired.Now, returning to the original problem, note that
![\[ (2n + 1)!\sum_{k = 0}^{n}(-1)^kB(n + k, n - k) = \]](http://latex.artofproblemsolving.com/9/e/7/9e7a3f58ef4cdb7596002c314f8705b8c9e41fed.png)
![\[ (2n + 1)!\sum_{k = 0}^{n}(-1)^k\int_{0}^{1}t^{n + k}(1 - t)^{n - k}dt = \]](http://latex.artofproblemsolving.com/3/2/f/32fe91a90eb86a20d8613ee61519785971846606.png)
![\[ (2n + 1)!\sum_{k = 0}^{n}(-1)^k\int_{0}^{1}\sum_{j = 0}^{n - k}\binom{n - k}{j}(-1)^jt^{n + k +j} dt = \]](http://latex.artofproblemsolving.com/8/8/a/88a5863fd484430a9520588b75880c001d23cd39.png)
![\[ (2n + 1)!\sum_{k = 0}^{n}(-1)^k\sum_{j = 0}^{n - k}\frac{(-1)^j}{n + k + j + 1}\binom{n - k}{j} = \]](http://latex.artofproblemsolving.com/0/e/e/0eed69d958c7893712cb46d0571f2bff3329d5bc.png)
![\[ (2n + 1)!\sum_{m = 0}^{n}\frac{(-1)^m}{n + m + 1}\sum_{j + k = m}\binom{n - k}{j} = \]](http://latex.artofproblemsolving.com/e/2/7/e27c1860b972087c4a562a76bbcd91939b4aa09e.png)
![\[ (2n + 1)!\sum_{m = 0}^{n}\frac{(-1)^m}{n + m + 1}\binom{n + 1}{m} = \]](http://latex.artofproblemsolving.com/4/d/e/4de4e4f37cb184ce86bb36a78c693c848587ec97.png)
![\[ (2n + 1)!\left(B(n, n + 1) + \frac{(-1)^n}{2(n + 1)}\right) = \]](http://latex.artofproblemsolving.com/0/3/e/03e5c0b7ac7d3567a432bb1733f6b2b64df976bb.png)
![\[ (2n + 1)!\left(\frac{n!(n + 1)!}{(2n + 2)!} + \frac{(-1)^n}{2n + 2}\right) \]](http://latex.artofproblemsolving.com/4/7/4/47439f1dc43b462b3f1410157ee450070b20323f.png)
And so we are done. Essentially this problem reduced to two applications of the binomial theorem on the integrand of the Beta function.
This post has been edited 1 time. Last edited by Wolstenholme, Oct 14, 2014, 11:45 PM
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