## Question 8(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $n=1$: $\begin{pmatrix}5&-8\\2&-3\end{pmatrix}^1=\begin{pmatrix}5&-8\\2&-3\end{pmatrix}$, $\begin{pmatrix}4\times1+1&-8(1)\\2\times1&1-4(1)\end{pmatrix}=\begin{pmatrix}5&-8\\2&-3\end{pmatrix}$, so true for $n=1$ | B1 | Must show substitution into RHS; minimum $\begin{pmatrix}4+1&-8\\2&1-4\end{pmatrix}=\begin{pmatrix}5&-8\\2&-3\end{pmatrix}$ |
| Assume true for $n=k$ so $\begin{pmatrix}5&-8\\2&-3\end{pmatrix}^k=\begin{pmatrix}4k+1&-8k\\2k&1-4k\end{pmatrix}$ | M1 | Makes assumption statement for some value of $n$ |
| $\begin{pmatrix}5&-8\\2&-3\end{pmatrix}^{k+1}=\begin{pmatrix}4k+1&-8k\\2k&1-4k\end{pmatrix}\begin{pmatrix}5&-8\\2&-3\end{pmatrix}$ or reverse order | M1 | Sets up correct multiplication statement either way |
| $\begin{pmatrix}4k+1&-8k\\2k&1-4k\end{pmatrix}\begin{pmatrix}5&-8\\2&-3\end{pmatrix}=\begin{pmatrix}5(4k+1)-16k&-8(4k+1)+24k\\10k+2(1-4k)&-16k-3(1-4k)\end{pmatrix}$ | A1 | Correct unsimplified matrix |
| $=\begin{pmatrix}4(k+1)+1&-8(k+1)\\2(k+1)&1-4(k+1)\end{pmatrix}$ | A1 | Correct simplified matrix with no errors, matching previously seen unsimplified form |
| If true for $n=k$ then true for $n=k+1$, true for $n=1$ so true for all positive integers $n$ | A1 | Correct conclusion conveying all four underlined ideas |

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## Question 8(ii) Way 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| When $n=1$: $4^{n+1}+5^{2n-1}=16+5=21$, so true for $n=1$ | B1 | Shows $f(1)=21$ |
| Assume true for $n=k$ so $4^{k+1}+5^{2k-1}$ is divisible by 21 | M1 | Assumption statement |
| $f(k+1)-f(k)=4^{k+2}+5^{2k+1}-4^{k+1}-5^{2k-1}$ | M1 | Attempts $f(k+1)-f(k)$ |
| $=4\times4^{k+1}+25\times5^{2k-1}-4^{k+1}-5^{2k-1}$ | | |
| $=3f(k)+21\times5^{2k-1}$ or e.g. $=24f(k)-21\times4^{k+1}$ | A1 | Correct expression for $f(k+1)-f(k)$ in terms of $f(k)$ |
| $f(k+1)=4f(k)+21\times5^{2k-1}$ or e.g. $f(k+1)=25f(k)-21\times4^{k+1}$ | A1 | Correct expression for $f(k+1)$ in terms of $f(k)$ |
| If true for $n=k$ then true for $n=k+1$, true for $n=1$ so true for all positive integers $n$ | A1 | Correct conclusion |

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## Question 8(ii) Way 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| When $n=1$: $4^{n+1}+5^{2n-1}=16+5=21$, so true for $n=1$ | B1 | |
| Assume true for $n=k$ so $4^{k+1}+5^{2k-1}$ is divisible by 21 | M1 | |
| $f(k+1)=4^{k+1+1}+5^{2(k+1)-1}$ | M1 | Attempts $f(k+1)$ |
| $f(k+1)=4\times4^{k+1}+5^{2k+1}=4\times4^{k+1}+4\times5^{2k-1}+25\times5^{2k-1}-4\times5^{2k-1}$ | A1 | Correctly obtains $4f(k)$ or $21\times5^{2k-1}$ |
| $f(k+1)=4f(k)+21\times5^{2k-1}$ | A1 | Correct expression for $f(k+1)$ in terms of $f(k)$ |
| If true for $n=k$ then true for $n=k+1$, true for $n=1$ so true for all positive integers $n$ | A1 | |

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## Question 8(ii) Way 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| When $n=1$: $f(1)=21$, true for $n=1$ | B1 | |
| Assume true for $n=k$ | M1 | |
| $f(k+1)-mf(k)=4^{k+2}+5^{2k+1}-m(4^{k+1}+5^{2k-1})$ | M1 | Attempts $f(k+1)-mf(k)$ |
| $=(4-m)4^{k+1}+5^{2k+1}-m\times5^{2k-1}=(4-m)(4^{k+1}+5^{2k-1})+21\times5^{2k-1}$ | A1 | Correct expression in terms of $f(k)$ |
| $=(4-m)(4^{k+1}+5^{2k-1})+21\times5^{2k-1}+mf(k)$ | A1 | Correct expression for $f(k+1)$ |
| If true for $n=k$ then true for $n=k+1$, true for $n=1$ so true for all positive integers $n$ | A1 | |

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## Question 8(ii) Way 4:

| Answer/Working | Mark | Guidance |
|---|---|---|
| When $n=1$: $f(1)=21$, true for $n=1$ | B1 | |
| Assume true for $n=k$ so $4^{k+1}+5^{2k-1}=21M$ | M1 | |
| $f(k+1)=4^{k+1+1}+5^{2(k+1)-1}$ | M1 | |
| $f(k+1)=4\times4^{k+1}+5^{2k+1}=4(21M-5^{2k-1})+5^{2k+1}$ | A1 | Correctly obtains $84M$ or $21\times5^{2k-1}$ |
| $f(k+1)=84M+21\times5^{2k-1}$ | A1 | Correct expression for $f(k+1)$ in terms of $M$ and $5^{2k-1}$ |
| If true for $n=k$ then true for $n=k+1$, true for $n=1$ so true for all positive integers $n$ | A1 | |

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