# Problem 3, USAMO 1975

A polynomial $p(x)$ of degree $n$ satisfies

$p(0) = 0, p(1) = \dfrac{1}{2}$,

$p(2) = \dfrac{2}{3},\cdots, p(n) =\dfrac{n}{n+1}$ . Find $p(n+1)$.

Solution 1 .

If $q(x)=(x+1)p(x)-x$ then $\deg q=n+1$ and $q(k)=0\,\forall k=\overline{0,n}$ , therefore $q(x)\equiv a\cdot x\cdot (x-1)\cdots (x-n)\, (1)$, from this, setting $x=-1$ we have $a=\dfrac{(-1)^{n+1}}{(n+1)!}\, (2)$.

By $(1)$ and $(2)$ we have $q(n+1)=(-1)^{n+1}$, therefore $p(n+1)=\dfrac{(-1)^{n+1}+n+1}{n+2}$.

Solution 2.

By Lagrange’s Interpolation Formula, we have $p(x)\equiv\sum_{i=0}^np(i)\cdot \prod_{j\not =i}\dfrac{x-j}{i-j}$.

From above, if setting $x=n+1$ then

$p(n+1)=\dfrac{1}{n+2}\cdot\sum_{i=0}^ni\cdot\binom{n+2}{i+1}\cdot (-1)^{n-i}$

$=\dfrac{1}{n+2}\cdot\sum_{k=1}^{n+1}(k-1)\cdot\binom{n+2}{k}\cdot (-1)^{n-k+1}$

$=-\dfrac{1}{n+2}\cdot (\sum_{k=0}^{n+2}k\cdot\binom{n+2}{k}\cdot (-1)^{n-k+2}-\\ (n+2)-\sum_{k=0}^{n+2}(-1)^{n-k+2}\cdot\binom{n+2}{k}+ (-1)^{n+2}+1)$

$=-\dfrac{1}{n+2}\cdot((\sum_{k=0}^{n+2}\binom{n+2}{k}\cdot x^k\cdot (-1)^{n-k+2})'_{x=1}-\\ (n+2)-(1-1)^{n+2}+(-1)^{n+2}+1)$

$=-\dfrac{1}{n+2} \cdot\left(((x-1)^{n+2})'_{x=1}-n-1+(-1)^{n+2}\right)$

$=\dfrac{(-1)^{n+1}+n+1}{n+2} \,\Box$.