Difference between revisions of "User talk:Baijiangchen"
Baijiangchen (talk | contribs) (→Sam's stuff) |
|||
Line 10: | Line 10: | ||
==Sam's stuff== | ==Sam's stuff== | ||
− | Let <math>W(n)=\sum_{ | + | Let <math>W(n)=\sum_{k=1}^{n}(\binom{x-1}{k-1}W(k-1)(n-k)!(2^{n-k}))</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_{ | + | <math>W(x+1)=\sum_{k=1}^{x+1}(\binom{x}{k-1}W(k-1)(x-k+1)!(2^{x-k+1}))</math> |
− | <math>=\sum_{ | + | <math>=\sum_{k=1}^{x+1}(\binom{x-1}{k-1}(\frac{x-k+1}{x})W(k-1)(x-i)!(x-k+1)(2^{x-k})(2))</math> |
− | <math>=\sum_{ | + | <math>=\sum_{k=1}^{x+1}(\binom{x-1}{k-1}W(i-1)(k-i)!(x-k+1)(2^{x-k})(2x))</math> |
− | <math>=2x\sum_{ | + | <math>=2x\sum_{k=1}^{x+1}(\binom{x-1}{k-1}W(i-1)(x-i)!(x-i+1)(2^{x-i}))</math> |
− | <math>=2x[W(n)=\sum_{ | + | <math>=2x[W(n)=\sum_{k=1}^{x}(\binom{x-1}{k-1}W(k-1)(x-k)!(2^{x-k}))+\binom{x-1}{x}W(x)(1)!(2)]</math> |
<math>=2x[(2x-1)!!+\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 01:29, 22 July 2012
If:
Then:
Sam's stuff
Let
Assume that for some integer ,
. We intend to show that
.
And this is where I'm stuck.
Return!