Standard +0.3 This is a straightforward proof by induction of a recurrence relation formula. It requires standard induction steps (base case, inductive hypothesis, inductive step) with routine algebraic manipulation. While it's a Further Maths topic, the mechanics are formulaic and require no novel insight—slightly easier than average overall but typical for FP1 induction questions.
A series of positive integers \(u_1, u_2, u_3, \ldots\) is defined by
$$u_1 = 6 \text{ and } u_{n+1} = 6u_n - 5, \text{ for } n \geq 1.$$
Prove by induction that \(u_n = 5 \times 6^{n-1} + 1\), for \(n \geq 1\).
[5]
A series of positive integers $u_1, u_2, u_3, \ldots$ is defined by
$$u_1 = 6 \text{ and } u_{n+1} = 6u_n - 5, \text{ for } n \geq 1.$$
Prove by induction that $u_n = 5 \times 6^{n-1} + 1$, for $n \geq 1$.
[5]
\hfill \mbox{\textit{Edexcel FP1 Q6 [5]}}