Prove that $$\sum_{x=0}^{n}(-1)^x\binom{n}{x}=0.$$
I think this is some sort of binomial theorem expansion but I am not sure how to start the proof. Does this make sense?
$$0=(1-1)^n=\sum_{x=0}^n\binom nx1^{n-x}(-1)^x$$
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Welcome. You can find how to format your posts here. People don't like to answer questions that are poorly formatted, do it for your own good. – Jakobian Sep 14 '18 at 23:41
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After it's been properly formatted with MathJax, it appears to be correct. – Bernard Sep 14 '18 at 23:45
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What is $n$? If $n=0$ is it not $1$? – max_zorn Sep 15 '18 at 00:41
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If you search using Approach0 or look in the frequent tab of the posts tagged binomial-coefficients+summation, it's very likely that you can find a few posts on this site about the same problem. Here is one of them: Proving $\sum_{k=0}^n(-1)^k\binom nk=0$. – Martin Sleziak Sep 18 '18 at 11:31
2 Answers
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$$(a+b)^n=\sum_{k=0}^{n}a^kb^{n-k}\binom{n}{k}$$
put $a=-1$ and $b=1$
$$(-1+1)^n=\sum_{k=0}^{n}(-1)^k1^{n-k}\binom{n}{k}$$ $$\sum_{k=0}^{n}(-1)^k1^{n-k}\binom{n}{k}=(1-1)^n=0$$
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Yes of course that’s the proof
$$0=(1-1)^n=\sum_{x=0}^n\binom nx1^{n-x}(-1)^x=\sum_{x=0}^n (-1)^x \binom n x $$
and similarly
$$2^n=(1+1)^n=\sum_{x=0}^n\binom nx1^{n-x}(1)^x=\sum_{x=0}^n \binom n x $$
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