Prove that $$\sum_{k=0}^n(-1)^k {{n}\choose{k}} = 0$$
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By the binomial theorem, we know that $$\sum_{k=0}^n(-1)^k {{n}\choose{k}} =\sum_{k=0}^n(-1)^k \times 1^{n-k} {{n}\choose{k}} =(1+(-1))^{n}= 0$$ We have the desired result. Also note that elementary sums involving binomial coefficients usually involves the binomial theorem.
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