## Question 10 – Aliter Way 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(n) = 2^{2n-1} + 3^{2n-1}$ is divisible by 5. | | |
| $f(1) = 2^1 + 3^1 = 5$ | B1 | Shows that $f(1) = 5$ |
| Assume for $n=k$, $f(k) = 2^{2k-1}+3^{2k-1}$ is divisible by 5 for $k \in \mathbb{Z}^+$ | | |
| $f(k+1) = 2^{2(k+1)-1} + 3^{2(k+1)-1} = 2^{2k+1}+3^{2k+1}$ | M1A1 | M1: Attempts $f(k+1)$. A1: Correct expression for $f(k+1)$ (can be unsimplified) |
| $= 4(2^{2k-1}) + 9(3^{2k-1})$ | M1 | Achieves an expression in $2^{2k-1}$ and $3^{2k-1}$ |
| $f(k+1) = 4(2^{2k-1}+3^{2k-1})+5(3^{2k-1})$ or $f(k+1) = 4f(k)+5(3^{2k-1})$ or $f(k+1) = 9f(k)-5(2^{2k-1})$ or $f(k+1) = 9(2^{2k-1}+3^{2k-1})-5(2^{2k-1})$ | A1 | Where $f(k+1)$ is correct and is clearly a multiple of 5. |
| If the result is true for $n=k$, then it is now true for $n=k+1$. As the result has been shown to be true for $n=1$, then the result is true for all $n$. | A1 cso | Correct conclusion at the end, at least as given, and all previous marks scored. |
| | **[6]** | |

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## Question 10 – Aliter Way 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(1) = 2^1 + 3^1 = 5$ | B1 | Shows that $f(1)=5$ |
| Assume for $n=k$, $f(k) = 2^{2k-1}+3^{2k-1}$ is divisible by 5 for $k \in \mathbb{Z}^+$ | | |
| $f(k+1)+f(k) = 2^{2(k+1)-1}+3^{2(k+1)-1}+2^{2k-1}+3^{2k-1}$ | M1A1 | M1: Attempts $f(k+1)+f(k)$. A1: Correct expression for $f(k+1)$ (can be unsimplified) |
| $= 2^{2k+1}+3^{2k+1}+2^{2k-1}+3^{2k-1}$ | | |
| $= 2^{2k-1+2}+3^{2k-1+2}+2^{2k-1}+3^{2k-1}$ | | |
| $= 4(2^{2k-1})+2^{2k-1}+9(3^{2k-1})+3^{2k-1}$ | M1 | Achieves an expression in $2^{2k-1}$ and $3^{2k-1}$ |
| $= 5(2^{2k-1})+10(3^{2k-1})$ | | |
| $= 5(2^{2k-1})+5(3^{2k-1})+5(3^{2k-1})$ | | |
| $= 5f(k)+5(3^{2k-1})$ | | |
| $\therefore f(k+1) = 4f(k)+5(3^{2k-1})$ or $4(2^{2k-1}+3^{2k-1})+5(3^{2k-1})$ | A1 | Where $f(k+1)$ is correct and is clearly a multiple of 5. |
| If the result is true for $n=k$, then it is now true for $n=k+1$. As the result has been shown to be true for $n=1$, then the result is true for all $n$. | A1 cso | Correct conclusion at the end, at least as given, and all previous marks scored. |
| | **[6]** | |

## Question 10 (Aliter, Way 4):

**Prove $f(n) = 2^{2n-1} + 3^{2n-1}$ is divisible by 5**

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| Working/Answer | Mark | Guidance |
|---|---|---|
| $f(1) = 2^1 + 3^1 = 5$ | B1 | Shows that $f(1) = 5$ |
| Assume that for $n = k$, $f(k) = 2^{2k-1} + 3^{2k-1}$ is divisible by 5 for $k \in \mathbb{Z}^+$ | — | Statement of assumption |
| $f(k+1) = f(k+1) + f(k) - f(k)$ | — | Strategy shown |
| $f(k+1) = 2^{2(k+1)-1} + 3^{2(k+1)-1} + 2^{2k-1} + 3^{2k-1} - (2^{2k-1} + 3^{2k-1})$ | M1A1 | M1: Attempts $f(k+1) + f(k) - f(k)$; A1: Correct expression for $f(k+1)$ (can be unsimplified) |
| $= 4(2^{2k-1}) + 9(3^{2k-1}) + 2^{2k-1} + 3^{2k-1} - (2^{2k-1} + 3^{2k-1})$ | M1 | Achieves an expression in $2^{2k-1}$ and $3^{2k-1}$ |
| $= 5(2^{2k-1}) + 10(3^{2k-1}) - (2^{2k-1} + 3^{2k-1})$ | — | — |
| $= 5\!\left((2^{2k-1}) + 2(3^{2k-1})\right) - (2^{2k-1} + 3^{2k-1})$ | — | — |
| $= 5\!\left((2^{2k-1}) + 2(3^{2k-1})\right) - f(k)$ or $5\!\left((2^{2k-1}) + 2(3^{2k-1})\right) - (2^{2k-1} + 3^{2k-1})$ | A1 | Where $f(k+1)$ is correct and is clearly a multiple of 5 |
| If the result is true for $n = k$, then it is now true for $n = k+1$. As the result has been shown to be true for $n = 1$, then the result is true for all $n$. | A1 cso | Correct conclusion **at the end**, at least as given, and all previous marks scored |

**Total: [6 marks]**