Mathematical Reflections 2017, Issue 1 - Problem U401

Problem:
Let $P$ be a polynomial of degree $n$ such that $P(k)=\dfrac{1}{k^2}$ for all $k=1,2,\ldots,n+1$. Determine $P(n+2)$.

Proposed by Dorin Andrica, Babe\c{s}-Bolyai University, Cluj-Napoca, Romania

Solution:
There exists a unique interpolating polynomial $P$ of degree $n$ such that $P(k)=\dfrac{1}{k^2}$ for all $k=1,2,\ldots,n+1$ and this is
$$P(x)=\sum_{k=1}^{n+1} \left(\prod_{\stackrel{1\leq j\leq n+1}{j\neq k}}\frac{x-j}{k-j}\right)\dfrac{1}{k^2}.$$
Observe that $$\renewcommand{\arraystretch}{2} \begin{array}{lll} \displaystyle \prod_{\stackrel{1\leq j\leq n+1}{j\neq k}}\frac{n+2-j}{k-j}&=&\displaystyle\prod_{j=1}^{k-1} \left(\dfrac{n+2-j}{k-j}\right)\prod_{j=k+1}^{n+1} \left(\dfrac{n+2-j}{k-j}\right)\\&=& \displaystyle \dfrac{(n+1)!}{(n-k+2)!(k-1)!}\cdot\dfrac{(n-k+1)!}{(-1)^{n-k+1}(n-k+1)!}\\&=& \displaystyle (-1)^{n-k+1}{n+1 \choose k-1}. \end{array}$$
So, $$P(n+2)=\sum_{k=1}^{n+1} (-1)^{n-k+1}{n+1 \choose k-1}\dfrac{1}{k^2}.$$