Prove by induction that $$\sum _ { n = 1 } ^ { N } n ^ { 3 } = \frac { 1 } { 4 } N ^ { 2 } ( N + 1 ) ^ { 2 }$$ Use this result, together with the formula for \(\sum _ { n = 1 } ^ { N } n ^ { 2 }\), to show that $$\sum _ { n = 1 } ^ { N } \left( 20 n ^ { 3 } + 36 n ^ { 2 } \right) = N ( N + 1 ) ( N + 3 ) ( 5 N + 2 ) .$$ Let $$S _ { N } = \sum _ { n = 1 } ^ { N } \left( 20 n ^ { 3 } + 36 n ^ { 2 } + \mu n \right)$$ Find the value of the constant \(\mu\) such that \(S _ { N }\) is of the form \(N ^ { 2 } ( N + 1 ) ( a N + b )\), where the constants \(a\) and \(b\) are to be determined. Show that, for this value of \(\mu\), $$5 + \frac { 22 } { N } < N ^ { - 4 } S _ { N } < 5 + \frac { 23 } { N }$$ for all \(N \geqslant 18\).