By considering \(( 2 r + 1 ) ^ { 3 } - ( 2 r - 1 ) ^ { 3 }\), use the method of differences to prove that
$$\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 1 )$$
Let \(S _ { n } = 1 ^ { 2 } + 3 \times 2 ^ { 2 } + 3 ^ { 2 } + 3 \times 4 ^ { 2 } + 5 ^ { 2 } + 3 \times 6 ^ { 2 } + \ldots + \left( 2 + ( - 1 ) ^ { n } \right) n ^ { 2 }\).
Show that \(\mathrm { S } _ { 2 \mathrm { n } } = \frac { 1 } { 3 } \mathrm { n } ( 2 \mathrm { n } + 1 ) ( \mathrm { an } + \mathrm { b } )\), where \(a\) and \(b\) are integers to be determined.
State the value of \(\lim _ { n \rightarrow \infty } \frac { S _ { 2 n } } { n ^ { 3 } }\).