## Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Let $P_n$ be the proposition that $5^n + 3$ is divisible by 4; $5^0 + 3 = 4 \Rightarrow P_0$ is true (allow $P_1$) | B1 | Some explanation of what $P_k$ being true means |
| Assume that $P_k$ is true for some non-negative integer $k$ | B1 | or e.g. $5^k + 3 = 4\alpha$ for 2nd B1 |
| $5^{k+1} + 3 = 5(4\alpha - 3) + 3$ | M1 | Alt method: Use $f(k+1) - f(k)$ **M1 A1** |
| $= 20\alpha - 12 = 4(5\alpha - 3)$ (or shows that $5^{k+1}+3 = 5 \cdot 5^k + 5 \cdot 3 - 4 \cdot 3 = 5(5^k+3) - 4 \cdot 3$) | A1 | |
| $P_0$ is true and $P_k \Rightarrow P_{k+1}$, hence $P_n$ is true for all non-negative integers $n$ | A1 | |
| **Total: 5** | | |

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