Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
The left hand side also gives the coefficient of $t$ as $n$ instead of $-n$, so you possibly want a $(-1)^i$ as well. – Matthew Pressland Jun 12 '12 at 12:04
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.