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I want to show, using the Binomial Theorem, that

$\binom{n}{0} - \binom{n}{1} + \binom{n}{2} + ... + \left ( -1 \right )^{n}\binom{n}{n} = 0$

By using the symmetry of $\binom{n}{0} = \binom{n}{n}$ it is easy for all even $n$, but how does one show it when n is odd? The middle combination of the sequence must then cancel out all the others, right? Meaning

$-\binom{^{n}}{\frac{n+1}{2}}+2\sum_{k=0}^{\frac{n-1}{2}} \binom{n}{k}-1^{k}=0$

Surely there is a better way.

Delta Foxtrot
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    Use the binomial theorem to evaluate $(1+(-1))^n$? – Rushabh Mehta Sep 11 '22 at 10:06
  • You probably mean $\binom{n}{0} - \binom{n}{1} + \binom{n}{2} + \cdots$ – Side note: the usual term for the opposite of “even” is “odd,” not “uneven.” – Martin R Sep 11 '22 at 10:19
  • @RushabhMehta Indeed, but that would be the reverse? I did identify you could write it like so, but then what? Should one evaluate the expression in the parenthesis and show that it is zero, as n is more than or equal to one? – Delta Foxtrot Sep 11 '22 at 10:28

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