Difference between revisions of "User talk:Baijiangchen"

Latest revision as of 21:23, 22 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 $W(n)=\sum_{k=1}^{n}(\binom{x-1}{k-1}W(k-1)(n-k)!(2^{n-k}))$

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

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

$=\sum_{k=1}^{x}(\binom{x-1}{k-1}(\frac{x}{x-k+1})W(k-1)(x-k)!(x-k+1)(2^{x-k})(2))+\binom{x}{x}W(x)(0)!(2^{0})$

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

$=2x(2x-1)!!+(2x-1)!!=(2x-1)!!(2x+1)=(2x+1)!!$

Q.E.D.

@@ Line 10: / Line 10: @@
 ==Sam's stuff==
-Let <math>W(n)=\sum_{i=1}^{n}(\binom{i-1}{n-1}W(i-1)(n-i)!(2^{n-i}))</math>
+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_{i=1}^{x+1}(\binom{i-1}{x}W(i-1)(x-i+1)!(2^{x-i+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_{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>=\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_{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\sum_{k=1}^{x}(\binom{x-1}{k-1}W(k-1)(x-k)!(2^{x-k}))+W(x)</math>
-<math>=2x[(2x-1)!!+\binom{x}{x-1}W(x)(1)!(2)]</math>
+<math>=2x(2x-1)!!+(2x-1)!!=(2x-1)!!(2x+1)=(2x+1)!!</math>
+Q.E.D.