Consider the sequence $S_0$ consisting of ones:
$$
1,1,1,1,1,1,\ldots
$$
Now compute the cumulative sum of this sequence, and call the resulting sequence $S_1$:
$$
1,2,3,4,5,6,\ldots
$$
Proceed iteratively to generate sequence $S_2$:
$$
1,3,6,10,15,21,\ldots
$$
then $S_3$:
$$
1,4,10,20,35,56,\ldots
$$
and so on.
It is well known that each sequence $S_k$ can be represented by a $k$-degree polynomial $P_k(n)$. For the above sequences the polynomials are
$$
P_0(n) = 1
$$
$$
P_1(n) = n
$$
$$
P_2(n) = \frac{n^2+n}{2}
$$
$$
P_3(n) = \frac{n^3+3n^2+2n}{6}
$$
My question: Is there a general formula for coefficients of the polynomial $P_k$? Or more generally, is there a formula to compute $S_k(n)$ as a function of $n$ and $k$? I mean a closed formula $S_k(n) = f(k,n)$ (not an iterative procedure such as the construction method I just described).
If that helps, actually I'm not interested in $S_k(n)$, but rather in the quotient $S_k(n)/S_k(n+1)$.
Best Answer
The numbers in sequence $S_k$ are the binomial coefficients $\binom{m}{k}$; the $n$-th term of $S_k$ is $\binom{n-1+k}{k} = \frac{(n-1+k)!}{(n-1)!k!}$. One can prove this by using that $$ \binom{i}{i} + \binom{i+1}{i} + \cdots + \binom{i+j}{i} = \binom{i+j+1}{i+1} $$ for any $i,j \geq 0$.
For $k=1$ we find $P_1(n) = \binom{n-1+1}{1} = \binom{n}{1} = n$, for $k=2$ we find $P_2(n) = \binom{n-1+2}{2} = \binom{n+1}{2} = \frac{n(n+1)}{2}$, for $k=3$ we find $P_3(n) = \binom{n-1+3}{3} = \binom{n+2}{3} = \frac{n(n+1)(n+2)}{6}$, etcetera.