**(i)**

| $u_{n+1} = 5u_n + 3$, $u_1 = 3$ and $u_n = \frac{3}{4}(5^n - 1)$ | | |
| $n = 1$; $u_1 = \frac{3}{4}(5^1 - 1) = \frac{3}{4}(4) = 3$ | B1 | Check that $u_n = \frac{3}{4}(5^n - 1)$ yields 3 when $n = 1$ |
| So $u_n$ is true when $n = 1$ | | |
| Assume that for $n = k$ that, $u_k = \frac{3}{4}(5^k - 1)$ is true for $k \in \mathbb{Z}^+$ | | |
| Then $u_{k+1} = 5u_k + 3$ | | |
| $= 5\left(\frac{3}{4}(5^k - 1)\right) + 3$ | M1 | Substituting $u_k = \frac{3}{4}(5^k - 1)$ into $u_{k+1} = 5u_k + 3$ |
| $= \frac{3}{4}(5)^{k+1} - \frac{15}{4} + 3$ | M1 | An attempt to multiply out in order to achieve $\pm\lambda(5^{k+1}) \pm$ constant |
| $= \frac{3}{4}(5)^{k+1} - \frac{3}{4}$ | | |
| $= \frac{3}{4}(5^{k+1} - 1)$ | A1 | $\frac{3}{4}(5^{k+1} - 1)$ |
| Therefore the general statement, $u_n = \frac{3}{4}(5^n - 1)$ 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 | A1 | True when $n = k+1$, then by induction the result is true for all positive integers |

**(ii)**

| $f(n) = 5(5^n) - 4n - 5$ is divisible by 16 | | |
| $f(1) = 5(5^1) - 4(1) - 5 = 16$, $\{\text{which is divisible by } 16\}$ | B1 | Shows that $f(1) = 16$ |
| $\{\therefore f(n) \text{ is divisible by } 16 \text{ when } n = 1.\}$ | | |
| Assume that for $n = k$, $f(k) = 5(5^k) - 4k - 5$ is divisible by 16 for $k \in \mathbb{Z}^+$ | | |
| $f(k+1) - f(k) = 5(5^{k+1}) - 4(k+1) - 5 - (5(5^k) - 4k - 5)$ | M1 | Applies $f(k+1) - f(k)$ |
| $= 5(5^{k+1}) - 4k - 4 - 5 - 5(5^k) + 4k + 5$ | A1 | Correct expression for $f(k+1) - f(k)$ |
| $= 25(5^k) - 4k - 4 - 5 - 5(5^k) + 4k + 5$ | | |
| $= 20(5^k) - 4$ | | |
| $= 4(5(5^k) - 4k - 5) + 16k + 20 - 4$ | | |
| $= 4(5(5^k) - 4k - 5) + 16k + 16$ | | |
| $= 4f(k) + 16(k+1)$ | | |
| $\therefore f(k+1) = 5f(k) + 16(k+1)$ | A1 | $f(k+1) = 5f(k) + 16(k+1)$ |
| $\{\therefore f(k+1) = 5f(k) + 16(k+1), \text{ which is divisible by } 16 \text{ as both } 5f(k) \text{ and } 16(k+1) \text{ are both divisible by } 16.\}$ | | |
| If the result is true for $n = k$, then it is now true for $n = k+1$. As the result has shown to be true for $n = 1$, then the result is true for all $n$ | A1 cso | Correct conclusion |

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# Aliter 7(b) Way 2:

| $\mathbf{M} = \mathbf{PQ}$ | | |
| $\begin{pmatrix} -6a & 7a \\ 2b & -b \end{pmatrix} = \begin{pmatrix} 3a & -2a \\ -b & 2b \end{pmatrix}\begin{pmatrix} q_1 & q_2 \\ q_3 & q_4 \end{pmatrix}$ | M1 | Writes down a relevant pair of simultaneous equations |
| $-6 = 3q_1 - 2q_3$, $7 = -3q_2 - 2q_4$, $-1 = -q_2 + 2q_4$ | | | Can be implied by later working |
| $= \begin{pmatrix} -2 & 3 \\ 0 & 1 \end{pmatrix}$ | A1, A1 | Two out of four elements correct; Correct matrix |

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# Aliter 10(ii) Way 2:

| $f(n) = 5(5^n) - 4n - 5$ is divisible by 16 | | |
| $f(1) = 5(5^1) - 4(1) - 5 = 16$, $\{\text{which is divisible by } 16\}$ | B1 | Shows that $f(1) = 16$ |
| $\{\therefore f(n) \text{ is divisible by } 16 \text{ when } n = 1.\}$ | | |
| Assume that for $n = k$, $f(k) = 5(5^k) - 4k - 5$ is divisible by 16 for $k \in \mathbb{Z}^+$ | | |
| $f(k+1) = 5(5^{k+1}) - 4(k+1) - 5$ | M1 | Applies $f(k+1)$ |
| $= 25(5^k) - 4k - 9$ | A1 | Correct expression for $f(k+1)$ |
| $= 5(5(5^k) - 4k - 5) + 20k + 25 - 4k - 9$ | | |
| $= 5(5(5^k) - 4k - 5) + 16(k+1)$ | | |
| $\therefore f(k+1) = 5f(k) + 16(k+1)$ | A1 | $f(k+1) = 5f(k) + 16(k+1)$ |
| $\{\therefore f(k+1) = 5f(k) + 16(k+1), \text{ which is divisible by } 16 \text{ as both } 5f(k) \text{ and } 16(k+1) \text{ are both divisible by } 16.\}$ | | |
| If the result is true for $n = k$, then it is now true for $n = k+1$. As the result has shown to be true for $n = 1$, then the result is true for all $n$ | A1 cso | Correct conclusion |