# Question 9:

## Way 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(1) = 8^1 - 2^1 = 6$ | B1 | Shows $f(1) = 6$ |
| $f(k+1) - f(k) = 8^{k+1} - 2^{k+1} - (8^k - 2^k)$ | M1 | Attempt $f(k+1) - f(k)$ |
| $= 8^k(8-1) + 2^k(1-2) = 7\times8^k - 2^k$ | | |
| $= 6\times8^k + 8^k - 2^k \left(= 6\times8^k + f(k)\right)$ | M1A1 | M1: attempt $f(k+1)-f(k)$ as multiple of 6; A1: rhs is a correct multiple of 6 |
| $f(k+1) = 6\times8^k + 2f(k)$ | A1 | Completes to $f(k+1)$ is a multiple of 6 |
| Conclusion: true for $n=k \Rightarrow$ true for $n=k+1$; true for $n=1$ so true for all $n \in \mathbb{Z}^+$ | A1cso | Do not award if $n$ defined incorrectly e.g. '$n$ is an integer' |

## Way 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(1) = 6$ | B1 | Shows $f(1) = 6$ |
| $f(k+1) = 8^{k+1} - 2^{k+1} = 8(8^k - 2^k + 2^k) - 2\cdot2^k$ | M1 | Attempts $f(k+1)$ in terms of $2^k$ and $8^k$ |
| $f(k+1) = 8^{k+1} - 2^{k+1} = 8(f(k) + 2^k) - 2\cdot2^k$ | M1A1 | M1: attempts $f(k+1)$ in terms of $f(k)$; A1: rhs correct and multiple of 6 |
| $f(k+1) = 8f(k) + 6\cdot2^k$ | A1 | Completes to $f(k+1)$ is a multiple of 6 |
| Conclusion as above | A1cso | |

## Way 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(1) = 6$ | B1 | Shows $f(1) = 6$ |
| $f(k+1) - 8f(k) = 8^{k+1} - 2^{k+1} - 8\cdot8^k + 8\cdot2^k$ | M1 | Attempt $f(k+1) - 8f(k)$; any multiple $m$ replacing 8 awards M1 |
| $f(k+1) - 8f(k) = 8^{k+1} - 8^{k+1} + 8\cdot2^k - 2\cdot2^k = 6\cdot2^k$ | M1A1 | M1: attempt $f(k+1)-f(k)$ as multiple of 6; A1: rhs is correct multiple of 6 |
| $f(k+1) = 8f(k) + 6\cdot2^k$ | A1 | Completes to $f(k+1)$ is a multiple of 6; general form $f(k+1) = 6\cdot8^k + (2-m)(8^k - 2^k)$ |
| Conclusion as above | A1cso | |