Alternating sum over bottom of binomial coefficient with top fixed equals 0
From specialfunctionswiki
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
