Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 4 / 7D / a / Solution

For $n\ge 2$, use the binomial coefficient identity $(n/k)\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)$. It reduces the left side to $\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)(k/n{)}^{m-1}$. Put $t=(k-1)/(n-1)$. Then $k/n=((n-1)t+1)/n$, and the binomial theorem givesThus the required coefficients areFor $m=1$ the sum has the single coefficient $a_{n,1,0}=1$. For $k=1$, the power $t^0$ denotes the constant monomial $1$.

The original PDF allows $n=1$, but its displayed ratio $(k-1)/(n-1)$ is then $0/0$. Literally that expression is undefined. The meaningful endpoint identity is obtained before normalization:This polynomial identity is valid at $n=k=1$, giving $1$ on both sides. The normalized formula and the Bernstein monomial recurrence below are used only for $n\ge 2$.