(New page: Using Binomial Theorem, <math>(a+b)^n=\binom{n}{0}a^n+ \binom n 1 a^{n-1} b+...+\binom{n}{n}b^n</math>. <math>\binom{n}{0}+ \binom{n}{1}+...+\binom{n}{n}=(1+1)^n=2^n</math>)
 
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Using Binomial Theorem, <math>(a+b)^n=\binom{n}{0}a^n+ \binom n 1 a^{n-1} b+...+\binom{n}{n}b^n</math>.
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<math>\binom{n}{0}+ \binom{n}{1}+...+\binom{n}{n}=(1+1)^n=2^n</math>
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Revision as of 08:52, 7 September 2008

Alumni Liaison

Recent Math PhD now doing a post-doctorate at UC Riverside.

Kuei-Nuan Lin