The equation
$$x ^ { 3 } + 2 x ^ { 2 } + x - 100000 = 0$$
has one real root. Taking \(x _ { 1 } = 50\) as a first approximation to this root, use the Newton-Raphson method to find a second approximation, \(x _ { 2 }\), to the root.
Given that \(S _ { n } = \sum _ { r = 1 } ^ { n } r ( 3 r + 1 )\), use the formulae for \(\sum _ { r = 1 } ^ { n } r ^ { 2 }\) and \(\sum _ { r = 1 } ^ { n } r\) to show that
$$S _ { n } = n ( n + 1 ) ^ { 2 }$$
The lowest integer \(n\) for which \(S _ { n } > 100000\) is denoted by \(N\).
Show that
$$N ^ { 3 } + 2 N ^ { 2 } + N - 100000 > 0$$