Past exam of the options II course of the University of Oxford / 2012 / P / P4 / a / Solution

Subtract the two finite sums. In the shifted sum replace the summation index $k$ by $k-1$. For every interior index the new coefficient isby Pascal's identity; the two endpoint coefficients agree as well. ThereforeFor a degree-$d$ polynomial with leading coefficient $c\ne 0$, the backward difference has leading termSince $h>0$, this coefficient is nonzero for $d\ge 1$. Starting with $Q_{(0)}=P$, iteration gives degree $n-r$ and leading coefficient $c{h}^{r}n!/(n-r)!$ for $0\le r\le n$. The next difference of the resulting constant is zero: