Past exam of the mathematics course of the University of Cambridge / 2004 / ia / Paper 4 / 2E / Solution

For the odd-square sum, the initial case $n=1$ gives $1=(4-1)/3$. Assume the formula at $n$. Adding the next square givesThe initial case and this induction step prove $\boxed{\sum _{j=1}^n(2j-1{)}^2=(4n^3-n)/3}$ for every positive integer $n$.

For the divisibility assertion, $1^3+5\cdot 1=6$. The change when $n$ is replaced by $n+1$ isOne of the consecutive integers $n,n+1$ is even, so the displayed change is divisible by $6$. If $n^3+5n$ is divisible by $6$, its successor is too. Thus induction proves $\boxed{6\mid n^3+5n}$ for every $n\ge 1$.