Proving $\sum_{k=0}^n\dfrac{x_k^{n+1}}{\prod_{j\neq k}(x_k-x_j)}=\sum_{k=0}^nx_k$

In Problems from the book by Andreescu, there's the following problem :

Let $x_0,\ldots,x_n$ be distinct complex numbers.

1. Prove $\displaystyle \sum_{k=0}^n\dfrac{x_k^{n+1}}{\prod_{j\neq k}(x_k-x_j)}=\sum_{k=0}^nx_k$

2. Evaluate $\displaystyle \sum_{k=0}^n\dfrac{x_k^{n+2}}{\prod_{j\neq k}(x_k-x_j)}$

My try:

1. Set $\Pi= \prod_{i=0}^nx_i$ and $f:x\to \dfrac{(-1)^n}{\Pi}x^{n+2}$

• Setting $P$ as the Lagrange polynomial for the function $f$ and points $x_0,\ldots,x_n$ yields $$P(X)=\sum_{k=0}^n \dfrac{(-1)^n}{\Pi}x_k^{n+2} \prod_{j\neq k }\frac{X-x_j}{x_k-x_j}$$ and $$P(0)= \sum_{k=0}^n\dfrac{x_k^{n+1}}{\prod_{j\neq k}(x_k-x_j)}$$

It remains to prove that $P(0)=\sum_{k=0}^nx_k$, which would be true if hypothetically $P=\sum_{0}^nx_kX^k$

I don't know if indeed $P=\sum_{0}^nx_kX^k$ ...

• Besides, noting that $f$ is a polynomial with degree $n+2$, which matches with $P$ at $n+1$ points, there exists some $z$ such that $$f(X)-P(X)= \dfrac{(-1)^n}{\Pi}(X-x_0)\ldots(X-x_n)(X-z)$$

Hence $-P(0)=f(0)-P(0)=z$.

How to prove that $-z=\sum_{k=0}^nx_k$ ?

EDIT: Surprisingly (at least for me), this sum amounts to this one

$2.$ I haven't thought much about it, but I don't see any trivial connection with $1.$

• here is the answer to the first question given by....yourself :) – Petite Etincelle Aug 8 '14 at 11:38
• @LiuGang Yes, you're right. That is very odd, it didn't occur to me at anytime that both questions were related... Do you know how to deal with the second one ? (I think this is an overkill) – Gabriel Romon Aug 8 '14 at 12:32

Put $\displaystyle R(x)=\frac{x^{n+1}}{\prod_{k=0}^n (x-x_k)}$.
We have: $$R(x)=1+\sum_{k=0}^n \frac{b_k}{x-x_k}$$ with $$b_k=\lim_{x\to x_k}(x-x_k)R(x)=\frac{x_k^{n+1}}{\prod_{j\not = k}(x_k-x_j)}$$
Hence $$R(x)-1=\frac{x^{n+1}-\prod_{k=0}^n (x-x_k)}{\prod_{k=0}^n (x-x_k)}=\frac{x^{n}(\sum_{k=0}^n x_k)+...}{\prod_{k=0}^n (x-x_k)}=\sum_{k=0}^n \frac{b_k}{x-x_k}$$
Now you multiply this equality by $x$, and let $x\to +\infty$ and you are done.