Difference between revisions of "Alternating sum over bottom of binomial coefficient with top fixed equals 0"
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(Created page with "==Theorem== The following formula holds: $$\displaystyle\sum_{k=0}^n (-1)^k {n \choose k} = {n \choose 0} - {n \choose 1} + {n\choose 2} - \ldots \pm {n \choose n}=0,$$ where...") |
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$$\displaystyle\sum_{k=0}^n (-1)^k {n \choose k} = {n \choose 0} - {n \choose 1} + {n\choose 2} - \ldots \pm {n \choose n}=0,$$ | $$\displaystyle\sum_{k=0}^n (-1)^k {n \choose k} = {n \choose 0} - {n \choose 1} + {n\choose 2} - \ldots \pm {n \choose n}=0,$$ | ||
where the sign $\pm$ depends on whether $n$ is even or odd and ${n \choose k}$ denotes the [[binomial coefficient]]. | where the sign $\pm$ depends on whether $n$ is even or odd and ${n \choose k}$ denotes the [[binomial coefficient]]. | ||
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| + | ==Proof== | ||
==References== | ==References== | ||
* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Sum over bottom of binomial coefficient with top fixed equals 2^n|next=Finite sum of arithmetic progression}}: 3.1.7 | * {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Sum over bottom of binomial coefficient with top fixed equals 2^n|next=Finite sum of arithmetic progression}}: 3.1.7 | ||
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| + | [[Category:Theorem]] | ||
| + | [[Category:Unproven]] | ||
Latest revision as of 12:51, 17 September 2016
Theorem
The following formula holds: $$\displaystyle\sum_{k=0}^n (-1)^k {n \choose k} = {n \choose 0} - {n \choose 1} + {n\choose 2} - \ldots \pm {n \choose n}=0,$$ where the sign $\pm$ depends on whether $n$ is even or odd and ${n \choose k}$ denotes the binomial coefficient.
Proof
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): 3.1.7
