(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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Revision as of 09:51, 7 September 2008
Using Binomial Theorem, $ (a+b)^n=\binom{n}{0}a^n+ \binom n 1 a^{n-1} b+...+\binom{n}{n}b^n $. $ \binom{n}{0}+ \binom{n}{1}+...+\binom{n}{n}=(1+1)^n=2^n $
