Difference between revisions of "User talk:Baijiangchen"
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<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>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_{k=1}^{x | + | <math>=\sum_{k=1}^{x}(\binom{x-1}{k-1}(\frac{x}{x-k+1})W(k-1)(x-i)!(x-k+1)(2^{x-k})(2))+\binom{x}{x}W(x)(0)!(2^{0})</math> |
− | <math>=\sum_{k=1}^{x | + | <math>=2x\sum_{k=1}^{x}(\binom{x-1}{k-1}W(k-1)(x-k)!(2^{x-k}))+W(x)</math> |
− | <math>=2x | + | <math>=2x[(2x-1)!!+(2x-1)!!]=(2x-1)!!(2x+1)=(2x+1)!!</math> |
− | + | Q.E.D. | |
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Revision as of 01:35, 22 July 2012
If:
Then:
Sam's stuff
Let
Assume that for some integer ,
. We intend to show that
.
Q.E.D.