## Question 9(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $n=1$: LHS $= \begin{pmatrix} 3 & 0 \\ 6 & 1 \end{pmatrix}^1 = \begin{pmatrix} 3 & 0 \\ 6 & 1 \end{pmatrix}$, RHS $= \begin{pmatrix} 3^1 & 0 \\ 3(3^1-1) & 1 \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 6 & 1 \end{pmatrix}$, LHS = RHS, true for $n=1$ | B1 | Check to see that the result is true for $n=1$ |
| Assume true for $n=k$: $\begin{pmatrix} 3 & 0 \\ 6 & 1 \end{pmatrix}^k = \begin{pmatrix} 3^k & 0 \\ 3(3^k-1) & 1 \end{pmatrix}$ | | |
| $\begin{pmatrix} 3 & 0 \\ 6 & 1 \end{pmatrix}^{k+1} = \begin{pmatrix} 3^k & 0 \\ 3(3^k-1) & 1 \end{pmatrix}\begin{pmatrix} 3 & 0 \\ 6 & 1 \end{pmatrix}$ | M1 | $\begin{pmatrix} 3^k & 0 \\ 3(3^k-1) & 1 \end{pmatrix}$ by $\begin{pmatrix} 3 & 0 \\ 6 & 1 \end{pmatrix}$ |
| $= \begin{pmatrix} 3^{k+1}+0 & 0+0 \\ 9(3^k-1)+6 & 0+1 \end{pmatrix}$ | A1 | Correct unsimplified matrix with no errors seen |
| $= \begin{pmatrix} 3^{k+1} & 0 \\ 3(3(3^k)-1) & 1 \end{pmatrix} = \begin{pmatrix} 3^{k+1} & 0 \\ 3(3^{k+1}-1) & 1 \end{pmatrix}$ | dM1 | Manipulates so that $k \to k+1$ on at least one term |
| Correct result with no errors seen with some working between previous A1 | A1 | Correct result with no errors seen |
| If true for $n=k$ (1) then true for $n=k+1$ (2). True for $n=1$ (3), therefore true for all $n$ (4). **All 4 aspects needed** | A1 cso | Correct conclusion with all previous marks earned |

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## Question 9(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(1) = 7^{2-1}+5 = 7+5 = 12$, divisible by 12, so $f(n)$ divisible by 12 when $n=1$ | B1 | Shows that $f(1)=12$ |
| Assume for $n=k$: $f(k) = 7^{2k-1}+5$ is divisible by 12 for $k \in \mathbb{Z}^+$ | | |
| $f(k+1) = 7^{2(k+1)-1}+5$ | B1 | Correct unsimplified expression for $f(k+1)$ |
| $f(k+1) = 7^{2k+1}+5$ | | |
| $f(k+1)-f(k) = (7^{2k+1}+5)-(7^{2k-1}+5)$ | M1 | Applies $f(k+1)-f(k)$. No simplification necessary, condone missing brackets |
| $= 7^{2k+1}-7^{2k-1} = 7^{2k-1}(7^2-1)$ | M1 | Attempting to isolate $7^{2k-1}$ |
| $= 48(7^{2k-1})$ | A1cso | $48(7^{2k-1})$ |
| $f(k+1) = f(k)+48(7^{2k-1})$, both $f(k)$ and $48(7^{2k-1})$ divisible by 12. True for $n=k$ (1), true for $n=k+1$ (2), true for $n=1$ (3), true for all $n$ (4). **All 5 aspects needed** | A1 cso | Correct conclusion with no incorrect work. Don't condone missing brackets |

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## Question 9(b) Aliter Way 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(1)=12$, divisible by 12 | B1 | Shows $f(1)=12$ |
| $f(k+1) = 7^{2(k+1)-1}+5$ | B1 | Correct expression for $f(k+1)$ |
| $7^{2k+1}+5 = 49 \times 7^{2k-1}+5$ | M1 | Attempt to isolate $7^{2k-1}$ |
| $= 49\times(7^{2k-1}+5)-240$ | M1 | Attempt to isolate $7^{2k-1}+5$ |
| $f(k+1) = 49\times f(k)-240$ | A1 | Correct expression in terms of $f(k)$ |
| Both $f(k)$ and 240 divisible by 12, correct conclusion | A1 | Correct conclusion |

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## Question 9(b) Aliter Way 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(1)=12$, divisible by 12 | B1 | Shows $f(1)=12$ |
| $f(k+1) = 7^{2(k+1)-1}+5$ | B1 | Correct expression for $f(k+1)$ |
| $7^{2k+1}+5 = 7^2 \cdot 7^{2k-1}+5 = 49\times 7^{2k-1}+5$ | M1 | Attempt to isolate $7^{2k-1}$ |
| $= 49\times(12m-5)+5$ | M1 | Substitute for $m$ |
| $f(k+1) = 49\times 12m - 240$ | A1 | Correct expression in terms of $m$ |
| Both $49\times 12m$ and 240 divisible by 12, correct conclusion | A1 | Correct conclusion |

## Question 9(b) — Way 4:

| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $f(1) = 7^{2-1} + 5 = 7 + 5 = 12$ | B1 | Shows that $f(1) = 12$ |
| {which is divisible by 12}. $\therefore f(n)$ is divisible by 12 when $n=1$. | | |
| Assume that for $n = k$, $f(k) = 7^{2k-1} + 5$ is divisible by 12 for $k \in \mathbb{Z}^+$ | | |
| $f(k+1) + 35f(k) = 7^{2(k+1)-1} + 5 + 35(7^{2k-1}+5)$ | B1 | Correct expression for $f(k+1)$ |
| $f(k+1) + 35f(k) = 7^{2k+1} + 5 + 35(7^{2k-1}+5)$ | M1 | Add appropriate multiple of $f(k)$. For $7^{2k}$ this is likely to be 35 (119, 203,...). For $7^{2k-1}$ it is 11 (23, 35, 47,...) |
| giving, $7 \cdot 7^{2k} + 5 + 5 \cdot 7^{2k} + 175$ | M1 | Attempt to isolate $7^{2k}$ |
| $= 180 + 12 \times 7^{2k} = 12\left(15 + 7^{2k}\right)$ | A1 | Correct expression |
| $\therefore f(k+1) = 12\left(7^{2k}+15\right) - 35f(k)$. As both $f(k)$ and $12\left(7^{2k}+15\right)$ are divisible by 12 then so is $f(k+1)$. 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 | Correct conclusion |

**Total: (6 marks)**