Evaluation of $\sum_{i=1}^n\cfrac{\beta_i^k}{\prod_{j=1\\j\neq i}^n (\beta_j-\beta_i)}$ Trying to solve a problem I ended up working on a polynomial of degree $n-1$ in $t$, $p(t)=\sum_{k=0}^{n-1} a_k t^k$, whose coefficients are given by
$a_k=\sum_{i=1}^n\cfrac{\beta_i^k}{\prod_{j=1\\j\neq i}^n (\beta_j-\beta_i)}$
Here, $\beta_i$ are $n$ distinct positive and ordered real numbers, so that $0<\beta_1<\beta_2<\cdots<\beta_n$. What can I say about the coefficients $a_k$? I made some attempts and I think that
$a_k=\begin{cases}0 &k=0,\dots,n-2\\z\neq0 &k=n-1\end{cases}$
but I don't know if this is true, how to prove it and what $z$ would be. Do any of you know a solution or a reference for this problem?
Thanks
 A: Let $n=3$ and for ease of notation let the points be $a,b,c.$ Then under your conjecture, the expression using the power $k=2$ [which satisfies $k \le n-1]$, namely
$$\frac{a^2}{(a-b)(a-c)}+\frac{b^2}{(b-a)(b-c)}+\frac{c^2}{(c-a)(c-b)},$$
should be $0.$ However this expression is identically $1,$ after a check on a symbolic algebra calculator.
Added later:
Let $f(x)$ be any function, and let $S_f$ denote your expression for $a_k$ but with its numerator $\beta_i^k$ replaced by $f(\beta_i.)$
Also let $V_n$ denote the $n \times n$ Vandermonde matrix, whose $j$-th row is formed by the $(j-1)$th powers of the $\beta_i,$ for $1 \le i \le n.$
The determinant of $V_n$ is then the product of the factors $(\beta_j-\beta_i)$ for $1 \le i<j\le n.$
It can then be shown that he product $S_f \det (V_n)$ is the same as the determinant of the matrix obtained from $V_n$ when its last row is replaced by the row $[f(\beta_1),f(\beta_2),\cdots f(\beta_n.)].$
If we take $f(x)=x^{n-1}$ this shows that $a_{n-1}=1,$ since we now have the original Vandermonde matrix on both sides. And if $0<k<n-1$ we have on the right side the result of taking the Vandermonde matrix and replacing the last row with one of the rows above it, making that determinant zero.
