Re: IMO 1981,(ROU). shortlisted
One should be carful of thinking that the answer is somehow $\displaystyle P(n+1)=\binom{n+1}{n+1}^{-1}$; since $P$ is of degree $n$ has already been determined at $n+1$ points ($k=0,1,...,n$).
We try to build a polynomial identical to $P$. My idea was to find an expression $x_k$, $k=0,1,...,n$, such that $\displaystyle x_k=\begin{cases}1&x=k\\0&x\neq k\end{cases}$. Then simply $\displaystyle P(x)=x_0\binom{n+1}{0}^{-1}+x_1\binom{n+1}{1}^{-1}+\cdots+x_n\binom{n+1}{n}^{-1}$.
A good start is to look at the expression $f_k(x)=x\cdot(x-1)\cdots(x-(k-1))\widehat{(x-k)}(x-(k+1))\cdots(x-n)$, which is nonzero only when $x=k$. (the hat notation means 'without') So to gaurantee it's $1$ when $x=k$, we simply divide the above expression by $f_k(k)$, and therefore $\displaystyle x_k=\frac{f_k(x)}{f_k(k)}$.
Now we compute:
$\frac{f_k(n+1)}{f_k(k)}=\frac{(n+1)\cdot(n+1-1)\cdots(n+1-(k-1))\widehat{(n+1-k})(n+1-(k+1))\cdots(n+1-n)}{k\cdot(k-1)\cdots\widehat{(k-k)}(k-(k+1))\cdots(k-n)}=\frac{(n+1)!/(n+1-k)}{k!(-1)^{n-k}(n-k)!}=(-1)^{n-k}\binom{n+1}{k}$
Finally, $\displaystyle P(n+1)=\sum_{k=0}^{n}(n+1)_k\binom{n+1}{k}^{-1}=\sum_{k=0}^{n}\frac{f_k(n+1)}{f_k(k)}\binom{n+1}{k}^{-1}=\sum_{k=0}^{n}(-1)^{n-k}=\sum_{k=0}^{n}(-1)^k$.
So the answer is $0$ if $n$ is odd, $1$ if $n$ is even.
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