Challenging +1.8 This is a challenging Further Maths induction proof requiring two parts: establishing an inequality with exponential growth, then proving a ratio inequality. Students must handle the AM-GM inequality (a_n + 1/a_n ≥ 2), manipulate nested exponentials carefully, and use the inductive hypothesis creatively in the second part. The recursive definition with parameter λ and the function g(n) = λ^(n-1) adds conceptual complexity beyond standard induction exercises.
8 The sequence of real numbers \(a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots\) is such that \(a _ { 1 } = 1\) and
$$a _ { n + 1 } = \left( a _ { n } + \frac { 1 } { a _ { n } } \right) ^ { \lambda }$$
where \(\lambda\) is a constant greater than 1 . Prove by mathematical induction that, for \(n \geqslant 2\),
$$a _ { n } \geqslant 2 ^ { \mathrm { g } ( n ) }$$
where \(g ( n ) = \lambda ^ { n - 1 }\).
Prove also that, for \(n \geqslant 2 , \frac { a _ { n + 1 } } { a _ { n } } > 2 ^ { ( \lambda - 1 ) \mathrm { g } ( n ) }\).
8 The sequence of real numbers $a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots$ is such that $a _ { 1 } = 1$ and
$$a _ { n + 1 } = \left( a _ { n } + \frac { 1 } { a _ { n } } \right) ^ { \lambda }$$
where $\lambda$ is a constant greater than 1 . Prove by mathematical induction that, for $n \geqslant 2$,
$$a _ { n } \geqslant 2 ^ { \mathrm { g } ( n ) }$$
where $g ( n ) = \lambda ^ { n - 1 }$.
Prove also that, for $n \geqslant 2 , \frac { a _ { n + 1 } } { a _ { n } } > 2 ^ { ( \lambda - 1 ) \mathrm { g } ( n ) }$.
\hfill \mbox{\textit{CAIE FP1 2004 Q8 [9]}}