Moderate -0.3 This is a straightforward proof by induction with a given formula for a linear recurrence relation. The base case is trivial (u₁ = 1), and the inductive step requires only algebraic manipulation of the recurrence relation. While it tests understanding of induction structure, it's more routine than a typical A-level question since the formula is provided and no problem-solving or discovery is needed—just mechanical verification.
A sequence \(u_n\) is defined by \(u_{n+1} = 2u_n + 3\) and \(u_1 = 1\).
Prove by induction that \(u_n = 4 \times 2^{n-1} - 3\) for all positive integers \(n\). [5]
A sequence $u_n$ is defined by $u_{n+1} = 2u_n + 3$ and $u_1 = 1$.
Prove by induction that $u_n = 4 \times 2^{n-1} - 3$ for all positive integers $n$. [5]
\hfill \mbox{\textit{SPS SPS FM Pure 2023 Q2 [5]}}