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I need some help for showing, $$\displaystyle \sum_{\beta \leq \alpha} \binom{\alpha}{\beta}(-1)^{|\alpha-\beta|}=0,$$ where $\alpha, \beta\in\mathbb N_0^n$ and $N_0=\mathbb N\cup \{0\}$. Any help will be welcome, thanks..

Asaf Karagila
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PtF
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    Your notation seems to indicate that $\alpha$ and $\beta$ are n-tuples over the natural numbers and zero. I don't know what the binomial coefficient means in that case. If you meant that $\alpha$ and $\beta$ are natural numbers or zero, then the symmetry in Pascal's triangle should give you the result. Specifically the identity ${\alpha \choose \beta}$ = ${\alpha \choose \alpha -\beta}$ – Nitin Passa Oct 04 '13 at 00:40
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    @NitinPassa: that is standard notation. For a multi-index the "binomial coefficient" is defined as usual ${\alpha \choose \beta} = \alpha ! / \beta! (\alpha-\beta)!$ where $\alpha! = \alpha_1!\alpha_2! \cdots \alpha_n!$. – Willie Wong Oct 04 '13 at 07:46

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Hint: apply the multi-binomial theorem.

Let me show you how to do it for $n = 1$, and you can easily generalise.

When $n = 1$ so $\alpha,\beta$ are natural numbers, we have that (assuming $\alpha \geq 1$)

$$ 0 = 0^\alpha = (1 - 1)^\alpha = \sum_{\beta \leq \alpha} {\alpha \choose\beta} 1^\beta (-1)^{(\alpha -\beta)} $$

is precisely what you want. In your case, you need to find a vector $x$ such that $x^\beta (-x)^{\alpha - \beta} = (-1)^{|\alpha - \beta|}$ and you can argue exactly the same way.

Note that the statement in your post is false for $|\alpha| = 0$. In that case the left hand side has just one term

$$ { \vec{0} \choose \vec{0}} (-1)^{0} = 1 \neq 0 $$

where $\vec{0}$ is the multi-index $(0,0,\ldots,0)$.

Willie Wong
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