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The identity $$\sum _{k=0}^n (-1)^k {n \choose k} = 0$$

For all $\mathbb{N}$ can be obtained from setting $a=-1$ and $b=1$ in the binomial thrm $(a+b)^n = \sum_{k=0}^n {n \choose k} a^k b ^{n-k}$. However, the former identity doesn't hold when $n=0$, as we have that the LHS thereof is $1$. However, the binomial theorem does hold when $n =0$, and because the binomial theorem implies this identity, it seems logically that this identity should hold when $n=0$, but it doesn't. Why is this?

N. F. Taussig
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Princess Mia
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    Why would $0^0$ be $0$? – WimC Sep 14 '23 at 18:50
  • Why do you think that "identity" should hold for $n=0$ as well? Recall that in the context of combinatorics, we have that $0^0 = 1$, and indeed our sum here gives a sum of $1$ as well, just as we should have from $(1-1)^0$ – JMoravitz Sep 14 '23 at 18:50
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    The correct identity should have read $\sum\limits_{k=0}^n(-1)^k\binom{n}{k}=0^n$ which in the event that $n\neq 0$ can indeed be simplified to just $0$, but in the event that $n=0$ the identity correctly gives the answer of $1$. Your proposed identity made this invalid simplification without including a disclaimer. – JMoravitz Sep 14 '23 at 18:52

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There is no issue here. When you take $(x+y)^n=\sum_k \binom nk x^ky^{n-k}$ and set $x=1,y=-1$, you get $$ 0^n=\sum_{k=0}^n \binom nk (-1)^k $$ This is different from what you had; you wrote $0$ on the LHS, when it should have been $0^n$. Note that, in the context of combinatorics, and more generally in the context of integer-only arithmetic, the commonly adopted convention is $$ 0^0=1 $$ Therefore, $0$ is not an equivalent expression for $0^n$, because they differ when $n=0$.

The phenomenon you are observing is one of the very reasons why we define $0^0=1$; if we did not define $0^0$ this way, then the binomial theorem would not hold when $n=0$. For more reasons in favor of the $0^0=1$ convention for combinatorics and integer math, see Zero to the zero power – is $0^0=1$?.

Mike Earnest
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