## Problem 3

A polynomial $p(x)$ of degree $n$ satisfies $p(0)=0, p(1)=1 / 2, p(2)=2 / 3, \ldots, p(n)=n /(n+1)$. Find $p(n+1)$.


## Solution

The polynomial $(\mathrm{x}+1) \mathrm{p}(\mathrm{x})-\mathrm{x}$ has degree $\mathrm{n}+1$ and zeros at $0,1,2, \ldots, \mathrm{n}$, so it must be $\mathrm{k} \mathrm{x}(\mathrm{x}-1)(\mathrm{x}-2) \ldots(\mathrm{x}$ $-n)$. Also it has value 1 at $x=-1$, so $k(n+1)!(-1)^{n+1}$. Hence $(n+2) p(n+1)-(n+1)=(-1)^{n+1}$. So for $n$ odd, $p(n+1)=1$, and for $n$ even, $p(n+1)=n /(n+2)$.

