- Show that, for any value of the real constant \(b\), the equation \(x ^ { 3 } - ( b + 1 ) x + b = 0\) has \(x = 1\) as a solution.
Find all values of \(b\) for which this equation has exactly two real solutions
\section*{11. In the question you must show detailed reasoning}
Given that the coefficients of \(x , x ^ { 2 }\) and \(x ^ { 4 }\) in the expansion of \(( 1 + k x ) ^ { n }\) are the consecutive terms of a geometric series, where \(n \geq 4\) and \(k\) is a positive constant
- Show that \(k = \frac { 6 ( n - 1 ) } { ( n - 2 ) ( n - 3 ) }\)
- For the case when \(k = \frac { 7 } { 5 }\), find the value of \(n\).
- Given that \(= \frac { 7 } { 5 } , n\) is a positive integer, and that the first term of the geometric series is the coefficient of \(x\), find the number of terms required for the sum of the geometric series to exceed \(1.12 \times 10 ^ { 12 }\).
\section*{12. In the question you must show detailed reasoning}
Given that \(\log _ { a } x = \frac { \log _ { b } x } { \log _ { b } a }\) show that the sum of the infinite series, where \(n = 0,1,2 \ldots\),
$$\log _ { 2 } e - \log _ { 4 } e + \log _ { 16 } e - \cdots + ( - 1 ) ^ { n } \log _ { 2 ^ { 2 } } n e + \cdots$$
is
$$\frac { 1 } { \ln ( 2 \sqrt { 2 } ) }$$
\section*{Advanced GCE (H245)}
\section*{Further Mathematics A}
\section*{Formulae Booklet}
\section*{Pure Mathematics}
\section*{Arithmetic series}
\(S _ { n } = \frac { 1 } { 2 } n ( a + l ) = \frac { 1 } { 2 } n \{ 2 a + ( n - 1 ) d \}\)
\section*{Geometric series}
\(S _ { n } = \frac { a \left( 1 - r ^ { n } \right) } { 1 - r }\)
\(S _ { \infty } = \frac { a } { 1 - r }\) for \(| r | < 1\)
\section*{Binomial series}
\(( a + b ) ^ { n } = a ^ { n } + { } ^ { n } \mathrm { C } _ { 1 } a ^ { n - 1 } b + { } ^ { n } \mathrm { C } _ { 2 } a ^ { n - 2 } b ^ { 2 } + \ldots + { } ^ { n } \mathrm { C } _ { r } a ^ { n - r } b ^ { r } + \ldots + b ^ { n } \quad ( n \in \mathbb { N } )\),
where \({ } ^ { n } \mathrm { C } _ { r } = { } _ { n } \mathrm { C } _ { r } = \binom { n } { r } = \frac { n ! } { r ! ( n - r ) ! }\)
$$( 1 + x ) ^ { n } = 1 + n x + \frac { n ( n - 1 ) } { 2 ! } x ^ { 2 } + \ldots + \frac { n ( n - 1 ) \ldots ( n - r + 1 ) } { r ! } x ^ { r } + \ldots \quad ( | x | < 1 , n \in \mathbb { R } )$$