Problem from Indian Statistical Institute: B.Stat. (Hons.)

Let $P(x)$ be a polynomial of degree $11$ such that \( P(x) = \frac{1}{x+1}\), for \( x = 0,1,2,\dots,11.\) $$$$ Find the value of $P(12)$ $$$$ Solution: Let \( f(x) = (x+1)P(x)-1\), clearly $f(x)$ is a polynomial of degree 12. $$$$ Now for \( x \in \{0,1,2,\dots,11\}\), \(f(x) = (x+1)P(x)-1=\frac{x+1}{x+1}-1=1-1=0\) $$$$ This shows that $f(x)$ vanishes at the points \( x = 0,1,2,\dots,11.\) $$$$ $f(x)$ being of degree 12, the above statement assures that \( x = 0,1,2,\dots,11.\) are the possible roots of $f(x)$ $$$$ Therefore \(f(x)=(x+1)P(x)-1=a_{0}(x-0)(x-1)(x-2)\dots(x-11)\) $$$$ Letting $x=-1$ in the above equality, we have \(-1=a_{0}(-1)(-2)(-3)\dots(-12) \Rightarrow a_{0} = \frac{-1}{12!}\) $$$$ Therefore \(f(x)=(x+1)P(x)-1=\frac{-1}{12!}(x-0)(x-1)(x-2)\dots(x-11)\) $$$$ Now, letting $x=12$ we have \(13P(12)-1=\frac{-1}{12!}(12)(11)(10)\dots(1)=\frac{-12!}{12!}=-1\) $$$$ \(\Rightarrow 13P(12)-1=-1 \Rightarrow P(12) = 0 \)