Difference between revisions of "Sum over bottom of binomial coefficient with top fixed equals 2^n"
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| − | * {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Binomial coefficient (n choose n) equals 1|next=Alternating sum over bottom binomial coefficient with top fixed equals 0}}: 3.1.6 | + | * {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Binomial coefficient (n choose n) equals 1|next=Alternating sum over bottom of binomial coefficient with top fixed equals 0}}: 3.1.6 |
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| + | [[Category:Theorem]] | ||
| + | [[Category:Unproven]] | ||
Latest revision as of 12:51, 17 September 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
