tag:blogger.com,1999:blog-8883034.post110570245635212402..comments2024-08-24T00:41:35.396+02:00Comments on atdotde: How to sum binomialsRoberthttp://www.blogger.com/profile/06634377111195468947noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-8883034.post-1138484610359863352006-01-28T22:43:00.000+01:002006-01-28T22:43:00.000+01:00(1/2)((1+x)^(2n)+(1-x)^(2n))setting x=1 gives the ...(1/2)((1+x)^(2n)+(1-x)^(2n))<BR/><BR/>setting x=1 gives the proof.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8883034.post-1107723007270251642005-02-06T21:50:00.000+01:002005-02-06T21:50:00.000+01:00Hmm, this time I was able to log on!
Anyway, ther...Hmm, this time I was able to log on!<br /><br />Anyway, there's a tricky step in the proof I just gave.<br />The trick is to show that precisely half the subsets of<br />any finite set have an even number of elements.<br /><br />While plausible (and true), your students may still have<br />some fun showing it.John Baezhttps://www.blogger.com/profile/11573268162105600948noreply@blogger.comtag:blogger.com,1999:blog-8883034.post-1107722081143602482005-02-06T21:34:00.000+01:002005-02-06T21:34:00.000+01:00You ask why the sum of 2n choose 2k from k = 0 to ...You ask why the sum of 2n choose 2k from k = 0 to <br />k = n is 4^n/2. <br /><br />Well, this is the number of subsets of even size of<br />a 2n-element set. <br /><br />The number of subsets of a 2n-element set is 4^n, <br />and half these are even. That does it.<br /><br />By the way, I'm John Baez. I'm only posting <br />"anonymously" because I couldn't manage to log in<br />on that blogger website. I gave them a username, <br />password, my name and an email address, and it all <br />seemed to work fine. <br /><br />But when I try to "sign in" to post something, I keep<br />getting looped around back to the same page where I'm<br />supposed to type my username and password. <br /><br />Any hints?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8883034.post-1105826456189456642005-01-15T23:00:00.000+01:002005-01-15T23:00:00.000+01:00You obviously have a lower standard of proof for y...You obviously have a lower standard of proof for yourself compared to your students... :)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8883034.post-1105731821920496032005-01-14T20:43:00.000+01:002005-01-14T20:43:00.000+01:00That's easy...
Sorry, I don't know how to display...That's easy...<br /><br />Sorry, I don't know how to display math equations on this blog.<br /><br />(a+b)^{2n}=\sum_{k=0}^{n}\begin{pmatrix}2n\\2k\end{pmatrix}a^{2n-2k}b^{2k}+\sum_{k=0}^{n-1}\begin{pmatrix}2n\\2k+1\end{pmatrix}a^{2n-2k-1}b^{2k+1}<br /><br />So,<br /><br />4^n=(1+1)^{2n}=\sum_{k=0}^{n}\begin{pmatrix}2n\\2k\end{pmatrix}+\sum_{k=0}^{n-1}\begin{pmatrix}2n\\2k+1\end{pmatrix}<br /><br />and<br /><br />0=(1-1)^{2n}=\sum_{k=0}^{n}\begin{pmatrix}2n\\2k\end{pmatrix}-\sum_{k=0}^{n-1}\begin{pmatrix}2n\\2k+1\end{pmatrix}<br /><br />So,<br /><br />\sum_{k=0}^{n}\begin{pmatrix}2n\\2k\end{pmatrix}=\sum_{k=0}^{n-1}\begin{pmatrix}2n\\2k+1\end{pmatrix}=4^n/2Anonymousnoreply@blogger.com