Difference between revisions of "User talk:Baijiangchen"
Baijiangchen (talk | contribs) (Created page with "If: <math>W(0):=1</math> <math>W(n):=\sum_{i=0}^{n-1}({n-1 \choose i}W(i)(x-i-1)!(2^{x-i-1})</math> Then: W(n)=(2n-1)!!") |
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If: | If: | ||
+ | |||
<math>W(0):=1</math> | <math>W(0):=1</math> | ||
− | <math>W(n):=\sum_{i=0}^{n-1}({n-1 \choose i}W(i)(x-i-1)!(2^{x-i-1})</math> | + | |
+ | <math>W(n):=\sum_{i=0}^{n-1}({n-1 \choose i}W(i)(x-i-1)!(2^{x-i-1}))</math> | ||
Then: | Then: | ||
− | W(n)=(2n-1)!! | + | |
+ | <math>W(n)=(2n-1)!!</math> | ||
+ | |||
+ | ==Sam's stuff== | ||
+ | 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>. | ||
+ | |||
+ | <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}(\binom{x-1}{k-1}(\frac{x}{x-k+1})W(k-1)(x-k)!(x-k+1)(2^{x-k})(2))+\binom{x}{x}W(x)(0)!(2^{0})</math> | ||
+ | |||
+ | <math>=2x\sum_{k=1}^{x}(\binom{x-1}{k-1}W(k-1)(x-k)!(2^{x-k}))+W(x)</math> | ||
+ | |||
+ | <math>=2x(2x-1)!!+(2x-1)!!=(2x-1)!!(2x+1)=(2x+1)!!</math> | ||
+ | |||
+ | Q.E.D. |
Latest revision as of 22:23, 22 July 2012
If:
Then:
Sam's stuff
Let
Assume that for some integer ,
. We intend to show that
.
Q.E.D.