| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2014 |
| Session | June |
| Marks | 6 |
| Topic | Proof by induction |
| Type | Prove divisibility |
| Difficulty | Standard +0.3 This is a standard divisibility proof by induction with straightforward algebra. The inductive step requires factoring out 121f(k) and showing the remainder term 7×4^n(121-16) is divisible by 13, which is routine manipulation. Slightly easier than average since the algebraic steps are mechanical once the standard approach is applied. |
| Spec | 4.01a Mathematical induction: construct proofs |
For $n = 1$, $\mathrm{f}(1) = 39\ (= 3 \times 13)$ which is divisible by 13 (so result true for $n = 1$) **B1**
Assume that $\mathrm{f}(k) = 13m$ (for some (positive) integer $m$) **M1**
Considering $\mathrm{f}(k+1) = 11^{2k+1} + 7 \times 4^{k+1}$ (must be simplified powers) **M1**
Expressing $\mathrm{f}(k+1)$ in terms of $\mathrm{f}(k)$
$\mathrm{f}(k+1) = 121\mathrm{f}(k) - 819 \times 4^k$ **or** $\mathrm{f}(k+1) = 4\mathrm{f}(k) + 117 \times 11^{2k-1}$ etc. **M1**
$= 13\{121m - 63 \times 4^k\}$ **or** $= 13\{4m + 9 \times 11^{2k-1}\}$ etc. **A1**
Explanation that $13 \mid \mathrm{f}(k) \Rightarrow 13 \mid \mathrm{f}(k+1)$ and $13 \mid \mathrm{f}(1) \Rightarrow$ general result **E1** [6]7 Let $\mathrm { f } ( n ) = 11 ^ { 2 n - 1 } + 7 \times 4 ^ { n }$. Prove by induction that $\mathrm { f } ( n )$ is divisible by 13 for all positive integers $n$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2014 Q7 [6]}}