The forward difference of a polynomial $P$ is $\Delta P(x)=P(x+1)-P(x)$, which is a new polynomial with degree reduced by one. Therefore, if we apply this operation three times we'll get a constant function, and we can work back up to get a value of $P(5)$. Practically, we create the following table of differences: $$\begin{array}{cccccc} \frac{1}{3} & \frac{1}{7} & & \frac{1}{13} & & \frac{1}{21} \\ \frac{-4}{21} & & \frac{-6}{91} & & \frac{-8}{273} & \\ & \frac{34}{273} & & \frac{10}{273} & & \end{array}$$ Then extend it to be the following table: So our answer is $\frac{-9}{273}=\frac{-3}{91}$
Let $P(x)$ be a polynomial of degree at most 3 such that $P(x)=\frac{1}{1+x+x^{2}}$ for $x=1,2,3,4$. What is $P(5) ?$