FP1 January 2011 Q9
9. A sequence of numbers \(u_1, u_2, u_3, u_4, \ldots\) is defined by \[u_{n+1} = 4u_n + 2, \quad u_1 = 2\]
Prove by induction that, for \(n \in \mathbb{Z}^+\), \[u_n = \frac{2}{3}(4^n - 1)\] (5)
| Scheme | Marks |
|---|---|
| \(u_{n+1} = 4u_n + 2,\ u_1 = 2\) and \(u_n = \tfrac{2}{3}(4^n - 1)\) | |
| \(n = 1;\ \ u_1 = \tfrac{2}{3}(4^1 - 1) = \tfrac{2}{3}(3) = 2\) So \(u_n\) is true when \(n = 1\). Check that \(u_n = \tfrac{2}{3}(4^n - 1)\) yields 2 when \(n = 1\). | B1 |
| Assume that for \(n = k\) that, \(u_k = \tfrac{2}{3}(4^k - 1)\) is true for \(k \in \mathbb{Z}^+\). Then \(u_{k+1} = 4u_k + 2\) | |
| \(= 4\left(\tfrac{2}{3}(4^k - 1)\right) + 2\) Substituting \(u_k = \tfrac{2}{3}(4^k - 1)\) into \(u_{n+1} = 4u_n + 2\). | M1 |
| \(= \tfrac{8}{3}(4)^k - \tfrac{8}{3} + 2\) An attempt to multiply out the brackets by 4 or \(\tfrac{8}{3}\) | M1 |
| \(= \tfrac{2}{3}(4)(4)^k - \tfrac{2}{3}\) \(= \tfrac{2}{3}4^{k+1} - \tfrac{2}{3}\) | |
| \(= \tfrac{2}{3}(4^{k+1} - 1)\) \(\tfrac{2}{3}(4^{k+1} - 1)\) | A1 |
| Therefore, the general statement, \(u_n = \tfrac{2}{3}(4^n - 1)\) is true when \(n = k + 1\). (As \(u_n\) is true for \(n = 1\),) then \(u_n\) is true for all positive integers by mathematical induction Require ‘True when n=1’, ‘Assume true when \(n=k\)’ and ‘True when \(n = k + 1\)’ then true for all \(n\) o.e. | A1 |
| (5) | |
| [5] |