Difference between revisions of "User talk:Baijiangchen"
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==Sam's stuff== | ==Sam's stuff== | ||
− | Let <math>W(n)=\sum_{i=1}^{n}(\binom{i-1}{n-1}W(i-1)( | + | Let <math>W(n)=\sum_{i=1}^{n}(\binom{i-1}{n-1}W(i-1)(n-i)!(2^{n-i}))</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>. | 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}^{ | + | <math>W(x+1)=\sum_{i=1}^{x+1}(\binom{i-1}{x}W(i-1)(x-i+1)!(2^{x-i+1}))</math> |
+ | <math>=\sum_{i=1}^{x+1}(\binom{i-1}{x-1}(\frac{x-i+1}{x})W(i-1)(x-i)!(x-i+1)(2^{x-i})(2))</math> | ||
+ | <math>=\sum_{i=1}^{x+1}(\binom{i-1}{x-1}W(i-1)(x-i)!(x-i+1)(2^{x-i})(2x))</math> | ||
+ | <math>=2x\sum_{i=1}^{x+1}(\binom{i-1}{x-1}W(i-1)(x-i)!(x-i+1)(2^{x-i}))</math> | ||
+ | <math>=2x[W(n)=\sum_{i=1}^{x}(\binom{i-1}{x-1}W(i-1)(x-i)!(2^{x-i}))+\binom{x}{x-1}W(x)(1)!(2)]</math> | ||
+ | <math>=2x[(2x-1)!!+\binom{x}{x-1}W(x)(1)!(2)]</math> |
Revision as of 00:37, 22 July 2012
If:
Then:
Sam's stuff
Let
Assume that for some integer ,
. We intend to show that
.