# Question 5:

## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| In first stage there is one square so $u_1 = 1$ | B1 | Explaining any two of the three aspects |
| Each square from $u_n$ to $u_{n+1}$ is replaced by 5 smaller squares, so $u_{n+1} = 5u_n$; but one square is removed so $u_{n+1} = 5u_n - 1$ | B1 | All three aspects explained |

## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| AE: $\lambda - 5 = 0 \Rightarrow \lambda = 5$ | M1 | Sets up and solves auxiliary equation |
| CF: $w_n = A \times 5^n$ | A1 | Correct complementary function |
| PS: try $v_n = k \Rightarrow k = 5k-1 \Rightarrow k = \frac{1}{4}$; so $u_n = A\times5^n + \frac{1}{4}$ | M1 | Correct form for particular solution, substitutes, combines with CF |
| $u_1 = 1 \Rightarrow 1 = A\times5 + \frac{1}{4} \Rightarrow A = \frac{3}{20}$ | M1 | Uses initial value to find constant |
| $u_n = \frac{3}{20}\times5^n + \frac{1}{4}$ or $u_n = \frac{3}{4}\times5^{n-1}+\frac{1}{4}$ | A1 | Correct solution |

## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Each square in stage $n$ has area $\frac{25}{9^{n-1}}$; total area $= \frac{25}{9^7}\times u_8 = \frac{25}{9^7}\times\left(\frac{3}{4}\times5^7+\frac{1}{4}\right)$ | M1 | Attempts scale factor with $u_8$; accept scaling by $25\times3^{-k}$ or $25\times9^{-k}$ where $k$ is 7, 8 or 9 |
| $\approx 0.3062$, accept awrt $0.31$ | A1 | Correct answer |

The image appears to be a blank page showing only the Pearson Education Limited company registration information at the bottom. There is no mark scheme content visible on this page to extract.