(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>. We have <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>. | 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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We have <math>\binom{n}{0}+ \binom{n}{1}+...+\binom{n}{n}=(1+1)^n=2^n</math> | We have <math>\binom{n}{0}+ \binom{n}{1}+...+\binom{n}{n}=(1+1)^n=2^n</math> | ||
Revision as of 09:53, 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 $.
We have $ \binom{n}{0}+ \binom{n}{1}+...+\binom{n}{n}=(1+1)^n=2^n $
