Edexcel FP1 — Question 6

Exam BoardEdexcel
ModuleFP1 (Further Pure Mathematics 1)
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve recurrence relation formula
DifficultyStandard +0.3 This is a straightforward proof by induction with a given formula for a recurrence relation. The base case is trivial (substituting n=1), and the inductive step requires only algebraic manipulation of the recurrence relation—no creative insight needed. While it's Further Maths content, this is a standard textbook exercise that's actually easier than many single-maths proof questions.
Spec1.04e Sequences: nth term and recurrence relations4.01a Mathematical induction: construct proofs

6. A series of positive integers \(u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots\) is defined by $$u _ { 1 } = 6 \text { and } u _ { n + 1 } = 6 u _ { n } - 5 , \text { for } n \geqslant 1 .$$ Prove by induction that \(u _ { n } = 5 \times 6 ^ { n - 1 } + 1\), for \(n \geqslant 1\).

Question 6:
Part (a) — Way 1
AnswerMarks Guidance
AnswerMarks Guidance
\(\mathbf{A} + \mathbf{B} = \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix}\)M1A1 M1: Correct attempt at matrix addition with 3 elements correct. A1: Correct matrix
\(2\mathbf{A} - \mathbf{B} = \begin{pmatrix} 5 & 1 \\ -2 & -1 \end{pmatrix}\)M1A1 M1: Correct attempt to double A and subtract B, 3 elements correct. A1: Correct matrix
\((\mathbf{A}+\mathbf{B})(2\mathbf{A}-\mathbf{B}) = \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix}\begin{pmatrix} 5 & 1 \\ -2 & -1 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ -7 & -2 \end{pmatrix}\)M1A1 M1: Correct method to multiply. A1: cao
Part (a) — Way 2
AnswerMarks Guidance
AnswerMarks Guidance
\((\mathbf{A}+\mathbf{B})(2\mathbf{A}-\mathbf{B}) = 2\mathbf{A}^2 + 2\mathbf{BA} - \mathbf{AB} - \mathbf{B}^2\)M1A1 M1: Expands brackets with at least 3 correct terms. A1: Correct expansion
\(\mathbf{A}^2 = \begin{pmatrix} 3 & 2 \\ -2 & -1 \end{pmatrix}\), \(\mathbf{BA} = \begin{pmatrix} -3 & -1 \\ -1 & 0 \end{pmatrix}\), \(\mathbf{AB} = \begin{pmatrix} -2 & 3 \\ 1 & -1 \end{pmatrix}\), \(\mathbf{B}^2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)M1A1 M1: Attempts \(\mathbf{A}^2\), \(\mathbf{B}^2\) and \(\mathbf{AB}\) or \(\mathbf{BA}\). A1: Correct matrices
\(2\mathbf{A}^2 + 2\mathbf{BA} - \mathbf{AB} - \mathbf{B}^2 = \begin{pmatrix} 1 & -1 \\ -7 & -2 \end{pmatrix}\)M1A1 M1: Substitutes into their expansion. A1: Correct matrix
Part (b) — Way 1
AnswerMarks Guidance
AnswerMarks Guidance
\(\mathbf{MC} = \mathbf{A} \Rightarrow \mathbf{C} = \mathbf{M}^{-1}\mathbf{A}\)B1 May be implied by later work
\(\mathbf{M}^{-1} = \frac{1}{-2-7}\begin{pmatrix} -2 & 1 \\ 7 & 1 \end{pmatrix}\)M1 An attempt at their \(\frac{1}{\det \mathbf{M}}\begin{pmatrix} -2 & 1 \\ 7 & 1 \end{pmatrix}\)
\(\mathbf{C} = \frac{1}{-2-7}\begin{pmatrix} -2 & 1 \\ 7 & 1 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ -1 & 0 \end{pmatrix}\)dM1 Correct order required and an attempt to multiply
\(\mathbf{C} = -\frac{1}{9}\begin{pmatrix} -5 & -2 \\ 13 & 7 \end{pmatrix}\)A1 oe
Part (b) — Way 2
AnswerMarks Guidance
AnswerMarks Guidance
\(\begin{pmatrix} 1 & -1 \\ -7 & -2 \end{pmatrix}\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ -1 & 0 \end{pmatrix}\)B1 Correct statement
\(a - c = 2,\ b - d = 1,\ -7a - 2c = -1,\ -7b - 2d = 0\)M1 Multiplies correctly to obtain 4 equations
\(a = \frac{5}{9},\ b = \frac{2}{9},\ c = -\frac{13}{9},\ d = -\frac{7}{9}\)M1A1 M1: Solves to obtain values for \(a, b, c\) and \(d\). A1: Correct values
# Question 6:

## Part (a) — Way 1

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{A} + \mathbf{B} = \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix}$ | M1A1 | M1: Correct attempt at matrix addition with 3 elements correct. A1: Correct matrix |
| $2\mathbf{A} - \mathbf{B} = \begin{pmatrix} 5 & 1 \\ -2 & -1 \end{pmatrix}$ | M1A1 | M1: Correct attempt to double **A** and subtract **B**, 3 elements correct. A1: Correct matrix |
| $(\mathbf{A}+\mathbf{B})(2\mathbf{A}-\mathbf{B}) = \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix}\begin{pmatrix} 5 & 1 \\ -2 & -1 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ -7 & -2 \end{pmatrix}$ | M1A1 | M1: Correct method to multiply. A1: cao |

## Part (a) — Way 2

| Answer | Marks | Guidance |
|--------|-------|----------|
| $(\mathbf{A}+\mathbf{B})(2\mathbf{A}-\mathbf{B}) = 2\mathbf{A}^2 + 2\mathbf{BA} - \mathbf{AB} - \mathbf{B}^2$ | M1A1 | M1: Expands brackets with at least 3 correct terms. A1: Correct expansion |
| $\mathbf{A}^2 = \begin{pmatrix} 3 & 2 \\ -2 & -1 \end{pmatrix}$, $\mathbf{BA} = \begin{pmatrix} -3 & -1 \\ -1 & 0 \end{pmatrix}$, $\mathbf{AB} = \begin{pmatrix} -2 & 3 \\ 1 & -1 \end{pmatrix}$, $\mathbf{B}^2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ | M1A1 | M1: Attempts $\mathbf{A}^2$, $\mathbf{B}^2$ and $\mathbf{AB}$ or $\mathbf{BA}$. A1: Correct matrices |
| $2\mathbf{A}^2 + 2\mathbf{BA} - \mathbf{AB} - \mathbf{B}^2 = \begin{pmatrix} 1 & -1 \\ -7 & -2 \end{pmatrix}$ | M1A1 | M1: Substitutes into their expansion. A1: Correct matrix |

## Part (b) — Way 1

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{MC} = \mathbf{A} \Rightarrow \mathbf{C} = \mathbf{M}^{-1}\mathbf{A}$ | B1 | May be implied by later work |
| $\mathbf{M}^{-1} = \frac{1}{-2-7}\begin{pmatrix} -2 & 1 \\ 7 & 1 \end{pmatrix}$ | M1 | An attempt at their $\frac{1}{\det \mathbf{M}}\begin{pmatrix} -2 & 1 \\ 7 & 1 \end{pmatrix}$ |
| $\mathbf{C} = \frac{1}{-2-7}\begin{pmatrix} -2 & 1 \\ 7 & 1 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ -1 & 0 \end{pmatrix}$ | dM1 | Correct order required and an attempt to multiply |
| $\mathbf{C} = -\frac{1}{9}\begin{pmatrix} -5 & -2 \\ 13 & 7 \end{pmatrix}$ | A1 | oe |

## Part (b) — Way 2

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix} 1 & -1 \\ -7 & -2 \end{pmatrix}\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ -1 & 0 \end{pmatrix}$ | B1 | Correct statement |
| $a - c = 2,\ b - d = 1,\ -7a - 2c = -1,\ -7b - 2d = 0$ | M1 | Multiplies correctly to obtain 4 equations |
| $a = \frac{5}{9},\ b = \frac{2}{9},\ c = -\frac{13}{9},\ d = -\frac{7}{9}$ | M1A1 | M1: Solves to obtain values for $a, b, c$ and $d$. A1: Correct values |

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6. A series of positive integers $u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots$ is defined by

$$u _ { 1 } = 6 \text { and } u _ { n + 1 } = 6 u _ { n } - 5 , \text { for } n \geqslant 1 .$$

Prove by induction that $u _ { n } = 5 \times 6 ^ { n - 1 } + 1$, for $n \geqslant 1$.\\

\hfill \mbox{\textit{Edexcel FP1  Q6}}