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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 $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood