Difference between revisions of "Sum over bottom of binomial coefficient with top fixed equals 2^n"
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Revision as of 02:49, 4 June 2016
Theorem
The following formula holds: $$\displaystyle\sum_{k=0}^n {n \choose k} = {n \choose 0} + {n \choose 1} + \ldots + {n \choose n} = 2^n,$$ where ${n \choose k}$ denotes the binomial coefficient.
Proof
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): 3.1.6