Question 9 - A-Level Maths

Edexcel FP1 2013 June — Question 9 10 marks

Exam Board	Edexcel
Module	FP1 (Further Pure Mathematics 1)
Year	2013
Session	June
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Proof by induction
Type	Prove recurrence relation formula
Difficulty	Standard +0.8 This is a two-part induction question from Further Maths FP1. Part (a) requires proving a recurrence relation formula involving exponential and linear terms, which is more complex than standard induction proofs. Part (b) involves matrix powers, requiring matrix multiplication in the inductive step. Both parts demand careful algebraic manipulation beyond routine A-level questions, but follow standard induction structure without requiring novel insights. The Further Maths context and multi-step algebraic complexity place this moderately above average difficulty.
Spec	4.01a Mathematical induction: construct proofs

9. (a) A sequence of numbers is defined by $$\begin{aligned} & u _ { 1 } = 8 \\ & u _ { n + 1 } = 4 u _ { n } - 9 n , \quad n \geqslant 1 \end{aligned}$$ Prove by induction that, for $n \in \mathbb { Z } ^ { + }$, $$u _ { n } = 4 ^ { n } + 3 n + 1$$ (b) Prove by induction that, for $m \in \mathbb { Z } ^ { + }$, $$\left( \begin{array} { l l } 3 & - 4 \\ 1 & - 1 \end{array} \right) ^ { m } = \left( \begin{array} { c c } 2 m + 1 & - 4 m \\ m & 1 - 2 m \end{array} \right)$$

Show mark scheme Show mark scheme source

Question 9:

Part (a):

Answer	Marks	Guidance
$u_1 = 8$ given; $n=1 \Rightarrow u_1 = 4^1 + 3(1) + 1 = 8$	B1	$4^1 + 3(1) + 1 = 8$ seen.

Assume true for $n = k$ so that $u_k = 4^k + 3k + 1$

Answer	Marks	Guidance
$u_{k+1} = 4(4^k + 3k + 1) - 9k$	M1	Substitute $u_k$ into $u_{k+1}$ as $u_{k+1} = 4u_k - 9k$
$= 4^{k+1} + 12k + 4 - 9k = 4^{k+1} + 3k + 4$	A1	Expression of the form $4^{k+1} + ak + b$
$= 4^{k+1} + 3(k+1) + 1$	A1	Correct completion to an expression in terms of $k+1$
If true for $n = k$ then true for $n = k+1$ and as true for $n = 1$ true for all $n$	A1 cso	Conclusion with all 4 underlined elements that can be seen anywhere in the solution; $n$ defined incorrectly award A0.

(5 marks)

Part (b):

$lhs = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}^1 = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}$

Answer	Marks	Guidance
$rhs = \begin{pmatrix} 2(1)+1 & -4(1) \\ 1 & 1-2(1) \end{pmatrix} = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}$	B1	Shows true for $m = 1$

Assume $\begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}^k = \begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix}$

Answer	Marks	Guidance
$\begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}^{k+1} = \begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix}\begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}$	M1	$\begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix}$ award M1
$= \begin{pmatrix} 6k+3-4k & -8k-4+4k \\ 3k+1-2k & -4k-1+2k \end{pmatrix}$	A1	Or equivalent $2\times2$ matrix. $\begin{pmatrix} 6k+3-4k & -12k-4+8k \\ 2k+1-k & -4k-1+2k \end{pmatrix}$ award A1 from above.

$= \begin{pmatrix} 2k+3 & -4k-4 \\ k+1 & -2k-1 \end{pmatrix}$

Answer	Marks	Guidance
$= \begin{pmatrix} 2(k+1)+1 & -4(k+1) \\ k+1 & 1-2(k+1) \end{pmatrix}$	A1	Correct completion to a matrix in terms of $k+1$
If true for $m = k$ then true for $m = k+1$ and as true for $m = 1$ true for all $m$	A1 cso	Conclusion with all 4 underlined elements that can be seen anywhere in the solution; $m$ defined incorrectly award A0.

(5 marks)

Total: 10 marks

# Question 9:

## Part (a):
$u_1 = 8$ given; $n=1 \Rightarrow u_1 = 4^1 + 3(1) + 1 = 8$ | B1 | $4^1 + 3(1) + 1 = 8$ seen.

Assume true for $n = k$ so that $u_k = 4^k + 3k + 1$

$u_{k+1} = 4(4^k + 3k + 1) - 9k$ | M1 | Substitute $u_k$ into $u_{k+1}$ as $u_{k+1} = 4u_k - 9k$

$= 4^{k+1} + 12k + 4 - 9k = 4^{k+1} + 3k + 4$ | A1 | Expression of the form $4^{k+1} + ak + b$

$= 4^{k+1} + 3(k+1) + 1$ | A1 | Correct completion to an expression in terms of $k+1$

If true for $n = k$ then true for $n = k+1$ and as true for $n = 1$ true for all $n$ | A1 cso | Conclusion with all 4 underlined elements that can be seen anywhere in the solution; $n$ defined incorrectly award A0.

**(5 marks)**

## Part (b):
$lhs = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}^1 = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}$

$rhs = \begin{pmatrix} 2(1)+1 & -4(1) \\ 1 & 1-2(1) \end{pmatrix} = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}$ | B1 | Shows true for $m = 1$

Assume $\begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}^k = \begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix}$

$\begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}^{k+1} = \begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix}\begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}$ | M1 | $\begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix}$ award M1

$= \begin{pmatrix} 6k+3-4k & -8k-4+4k \\ 3k+1-2k & -4k-1+2k \end{pmatrix}$ | A1 | Or equivalent $2\times2$ matrix. $\begin{pmatrix} 6k+3-4k & -12k-4+8k \\ 2k+1-k & -4k-1+2k \end{pmatrix}$ award A1 from above.

$= \begin{pmatrix} 2k+3 & -4k-4 \\ k+1 & -2k-1 \end{pmatrix}$

$= \begin{pmatrix} 2(k+1)+1 & -4(k+1) \\ k+1 & 1-2(k+1) \end{pmatrix}$ | A1 | Correct completion to a matrix in terms of $k+1$

If true for $m = k$ then true for $m = k+1$ and as true for $m = 1$ true for all $m$ | A1 cso | Conclusion with all 4 underlined elements that can be seen anywhere in the solution; $m$ defined incorrectly award A0.

**(5 marks)**

**Total: 10 marks**

Show LaTeX source

9. (a) A sequence of numbers is defined by

$$\begin{aligned}
& u _ { 1 } = 8 \\
& u _ { n + 1 } = 4 u _ { n } - 9 n , \quad n \geqslant 1
\end{aligned}$$

Prove by induction that, for $n \in \mathbb { Z } ^ { + }$,

$$u _ { n } = 4 ^ { n } + 3 n + 1$$

(b) Prove by induction that, for $m \in \mathbb { Z } ^ { + }$,

$$\left( \begin{array} { l l } 
3 & - 4 \\
1 & - 1
\end{array} \right) ^ { m } = \left( \begin{array} { c c } 
2 m + 1 & - 4 m \\
m & 1 - 2 m
\end{array} \right)$$

\hfill \mbox{\textit{Edexcel FP1 2013 Q9 [10]}}

This paper (9 questions)

View full paper

Q1 4 Q2 5 Q3 7 Q4 9 Q5 10 Q6 11 Q7 11 Q8 8 Q9 10

Answer	Marks	Guidance
\(u_{k+1} = 4(4^k + 3k + 1) - 9k\)	M1	Substitute \(u_k\) into \(u_{k+1}\) as \(u_{k+1} = 4u_k - 9k\)
\(= 4^{k+1} + 12k + 4 - 9k = 4^{k+1} + 3k + 4\)	A1	Expression of the form \(4^{k+1} + ak + b\)
\(= 4^{k+1} + 3(k+1) + 1\)	A1	Correct completion to an expression in terms of \(k+1\)
If true for \(n = k\) then true for \(n = k+1\) and as true for \(n = 1\) true for all \(n\)	A1 cso	Conclusion with all 4 underlined elements that can be seen anywhere in the solution; \(n\) defined incorrectly award A0.

Edexcel FP1 2013 June — Question 9 10 marks

Question 9:

Part (a):

Assume true for \(n = k\) so that \(u_k = 4^k + 3k + 1\)

(5 marks)

Part (b):

\(lhs = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}^1 = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\)

Assume \(\begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}^k = \begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix}\)

\(= \begin{pmatrix} 2k+3 & -4k-4 \\ k+1 & -2k-1 \end{pmatrix}\)

(5 marks)

Total: 10 marks

This paper (9 questions)

Answer	Marks	Guidance
\(\begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}^{k+1} = \begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix}\begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\)	M1	\(\begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix}\) award M1
\(= \begin{pmatrix} 6k+3-4k & -8k-4+4k \\ 3k+1-2k & -4k-1+2k \end{pmatrix}\)	A1	Or equivalent \(2\times2\) matrix. \(\begin{pmatrix} 6k+3-4k & -12k-4+8k \\ 2k+1-k & -4k-1+2k \end{pmatrix}\) award A1 from above.

Answer	Marks	Guidance
\(= \begin{pmatrix} 2(k+1)+1 & -4(k+1) \\ k+1 & 1-2(k+1) \end{pmatrix}\)	A1	Correct completion to a matrix in terms of \(k+1\)
If true for \(m = k\) then true for \(m = k+1\) and as true for \(m = 1\) true for all \(m\)	A1 cso	Conclusion with all 4 underlined elements that can be seen anywhere in the solution; \(m\) defined incorrectly award A0.