Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 11G / c / Solution

It suffices first to prove the elementary primorial bound for positive integers $n$, by strong induction. The case $n=1$ is immediate.

If $n=2k$, every prime $p$ with $k<p\le 2k$ divides $\left(\genfrac{}{}{0ex}{}{2k}{k}\right)$: the numerator $(2k)!$ contains $p$, whereas the two copies of $k!$ do not. Since these primes are distinct,The induction hypothesis and the binomial theorem give

If $n=2k+1$, part (b) shows that every prime $k+2\le p\le 2k+1$ divides $\left(\genfrac{}{}{0ex}{}{2k+1}{k+1}\right)$. HenceThe two central coefficients of $(1+1{)}^{2k+1}$ are equal, soand thereforeThis completes the induction. For real $X\ge 1$,