True for $k = 1$, since $1^5 - 1 = 0$ which is divisible by 5 | B1 | 2.1 |

Assume true for $n = k$, then $k^5 - k$ is divisible by 5 | M1 | 2.1 |

Then $(k+1)^5 - (k+1)$ | |

$= k^5 + 5k^4 + 10k^3 + 10k^2 + 5k + 1 - k - 1$ | |

$= \left(5k^4 + 10k^3 + 10k^2 + 5k\right) + k^5 - k$ | |

$= 5(k^4 + 2k^3 + 2k^2 + k) + k^5 - k$ | A1 | 2.5 |

So if divisible by 5 for $n = k$ then also for $n = k+1$ | A1 | 2.2a |

True for $n = 1$, so true generally | |

| [4] | |

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# Question 4 (i):

Find normal by vector product | M1 | 1.1a |

$\mathbf{n} = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \times \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 7 \\ -1 \\ -5 \end{pmatrix}$ | A1 | 1.1 |

Taking $\mathbf{b} = (7, 1, 6)$ and $\mathbf{a} = (2, 6, -2)$ | M1 | 2.1 |

$\mathbf{b} - \mathbf{a} = (5, -5, 8)$

Using distance formula: | A1 | 1.1 |

$(\mathbf{b}-\mathbf{a}) \cdot \mathbf{n} = \begin{pmatrix} 5 \\ -5 \\ 8 \end{pmatrix} \cdot \begin{pmatrix} 7 \\ -1 \\ -5 \end{pmatrix} = 0$ | |

So distance $= 0$ so they intersect. | |

**Alternative:**

Any point on $l_1$ is $(7 + 2\lambda, 1 - \lambda, 6 + 3\lambda)$ | M1, A1 |

Any point on $l_2$ is $(2 + \mu, 6 + 2\mu, -2 + \mu)$ | |

Equate any two: $\Rightarrow \lambda = -3, \mu = -1$

and the point is $(1, 4, -3)$ | |

| [4] | |

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# Question 4 (ii):

Use of normal vector | B1 | 1.1 |

$7x - y - 5z = c$ | M1 | 1.1 |

Through $a$ or $b$ | A1 | 1.1 |

$\Rightarrow c = 18$

$\Rightarrow 7x - y - 5z = 18$ | |

| [3] | |

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