Difference between revisions of "User talk:Baijiangchen"

Revision as of 23:31, 21 July 2012

If:

$W(0):=1$

$W(n):=\sum_{i=0}^{n-1}({n-1 \choose i}W(i)(x-i-1)!(2^{x-i-1}))$

Then:

$W(n)=(2n-1)!!$

Let $W(n)=\sum_{i=1}^{n}(\binom{i-1}{n-1}W(i-1)(x-i)!(2^{x-i}))$

Assume that for some integer $x$ , $W(x)=(2x-1)!!$ . We intend to show that $W(x+1)=(2(x+1)-1)!!=(2x+1)!!$ .

$W(x+1)=\sum_{i=1}^{x+1}(\binom{i-1}{x}W(i-1)(x-i+1)!(2^{x-i+1}))=\sum_{i=1}^{n}(\binom{i-1}{x-1}(\frac{x-i+1}{x})W(i-1)(x-i)!(x-i+1)(2^{x-i})(2))$

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	Assume that for some integer <math>x</math>, <math>W(x)=(2x-1)!!</math>. We intend to show that <math>W(x+1)=(2(x+1)-1)!!=(2x+1)!!</math>.		Assume that for some integer <math>x</math>, <math>W(x)=(2x-1)!!</math>. We intend to show that <math>W(x+1)=(2(x+1)-1)!!=(2x+1)!!</math>.
		+
		+	<math>W(x+1)=\sum_{i=1}^{x+1}(\binom{i-1}{x}W(i-1)(x-i+1)!(2^{x-i+1}))=\sum_{i=1}^{n}(\binom{i-1}{x-1}(\frac{x-i+1}{x})W(i-1)(x-i)!(x-i+1)(2^{x-i})(2))</math>