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.
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.
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.