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we know that hilbert seris of n- variables polynomial ring is $\Sigma_{i} \binom{n-1+i}{i}t^{i}$

But, I don't know $\Sigma_{i} \binom{n-1+i}{i}t^{i}=(1-t)^{-n}$.

I wonder to prove in detail.

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The left hand side also gives the coefficient of $t$ as $n$ instead of $-n$, so you possibly want a $(-1)^i$ as well. –  Matt Pressland Jun 12 '12 at 12:04
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up vote 2 down vote accepted

After the modifications in my comment, we can write the right hand side as $\frac{1}{(1 - t)^n} = (1 + t + t^2 + \ldots)^n$. Now the coefficient of $t^i$ in $(1 + t + t^2 + \ldots)^n$ is the number of ways to distribute $i$ identical objects to $n$ distinct containers, which is the coefficient of $t^i$ on the left hand side.

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