| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2010 |
| Session | January |
| Marks | 8 |
| 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 Further Maths question requiring (a) standard proof by induction on a recurrence relation with straightforward algebra, and (b) deriving a summation formula which requires either another induction or telescoping insight. While mechanically routine for FP2 students, it demands careful algebraic manipulation across multiple steps and combines two proof techniques, placing it moderately above average difficulty. |
| Spec | 4.01a Mathematical induction: construct proofs4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| Assume true for \(n = k\) | M1A1 | |
| \(u_{k+1} = 2(3 \times 2^{k-1} - 1) + 1\) | ||
| \(= 3 \times 2^k - 1\) | A1 | \(2^{(k-1)+1}\) not necessarily seen |
| True for \(n = 1\) shown | B1 | |
| Method of induction clearly expressed | E1 | 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum_{r=1}^n u_r = \sum_{r=1}^n 3 \times 2^{r-1} - n\) | M1 | |
| \(= 3(2^n - 1) - n\) | M1A1 | M1 for summation, ie recognition of a GP |
| \(= u_{n+1} - (n+2)\) | A1 | 3 marks |
### Part (a)
Assume true for $n = k$ | M1A1 |
$u_{k+1} = 2(3 \times 2^{k-1} - 1) + 1$ |
$= 3 \times 2^k - 1$ | A1 | $2^{(k-1)+1}$ not necessarily seen
True for $n = 1$ shown | B1 |
Method of induction clearly expressed | E1 | 5 marks | Provided all 4 previous marks earned
### Part (b)
$\sum_{r=1}^n u_r = \sum_{r=1}^n 3 \times 2^{r-1} - n$ | M1 |
$= 3(2^n - 1) - n$ | M1A1 | M1 for summation, ie recognition of a GP
$= u_{n+1} - (n+2)$ | A1 | 3 marks | AG7 The sequence $u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots$ is defined by
$$u _ { 1 } = 2 , \quad u _ { k + 1 } = 2 u _ { k } + 1$$
(a) Prove by induction that, for all $n \geqslant 1$,
$$u _ { n } = 3 \times 2 ^ { n - 1 } - 1$$
(b) Show that
$$\sum _ { r = 1 } ^ { n } u _ { r } = u _ { n + 1 } - ( n + 2 )$$
\hfill \mbox{\textit{AQA FP2 2010 Q7 [8]}}