Show that \(\left( n + \frac { 1 } { 2 } \right) ^ { 3 } - \left( n - \frac { 1 } { 2 } \right) ^ { 3 } \equiv 3 n ^ { 2 } + \frac { 1 } { 4 }\). Use this result to prove that \(\sum _ { n = 1 } ^ { N } n ^ { 2 } = \frac { 1 } { 6 } N ( N + 1 ) ( 2 N + 1 )\). The sums \(S , T\) and \(U\) are defined as follows: $$\begin{aligned} & S = 1 ^ { 2 } + 2 ^ { 2 } + 3 ^ { 2 } + 4 ^ { 2 } + \ldots + ( 2 N ) ^ { 2 } + ( 2 N + 1 ) ^ { 2 } ,
& T = 1 ^ { 2 } + 3 ^ { 2 } + 5 ^ { 2 } + 7 ^ { 2 } + \ldots + ( 2 N - 1 ) ^ { 2 } + ( 2 N + 1 ) ^ { 2 } ,
& U = 1 ^ { 2 } - 2 ^ { 2 } + 3 ^ { 2 } - 4 ^ { 2 } + \ldots - ( 2 N ) ^ { 2 } + ( 2 N + 1 ) ^ { 2 } . \end{aligned}$$ Find and simplify expressions in terms of \(N\) for each of \(S , T\) and \(U\). Hence

1. describe the behaviour of \(\frac { S } { T }\) as \(N \rightarrow \infty\),
2. prove that if \(\frac { S } { U }\) is an integer then \(\frac { T } { U }\) is an integer.