| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove summation with exponentials |
| Difficulty | Challenging +1.2 This is a structured induction proof with a helpful algebraic identity provided in part (a). While it involves factorials and exponentials making it non-trivial, the question guides students through the key algebraic manipulation needed for the inductive step. The mechanics are standard for FP2 induction, requiring careful algebra but no novel insight beyond applying the given result. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{1}{(k+2)!} - \frac{k+1}{(k+3)!} = \frac{k+3}{(k+3)!} - \frac{k+1}{(k+3)!}\) | M1 | Common denominator of \((k+3)!\) |
| \(= \frac{(k+3)-(k+1)}{(k+3)!} = \frac{2}{(k+3)!}\) | A1 | Completion to given result |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Base case: \(n=1\): LHS \(= \frac{1 \times 2}{3!} = \frac{2}{6} = \frac{1}{3}\); RHS \(= 1 - \frac{4}{3!} = 1 - \frac{2}{3} = \frac{1}{3}\) ✓ | B1 | Verify \(n=1\) |
| Assume \(\sum_{r=1}^{k} \frac{r \times 2^r}{(r+2)!} = 1 - \frac{2^{k+1}}{(k+2)!}\) | M1 | Statement of assumption for \(n=k\) |
| For \(n=k+1\): Add \(\frac{(k+1)2^{k+1}}{(k+3)!}\) to both sides | M1 | Attempt to add \((k+1)\)th term |
| \(= 1 - \frac{2^{k+1}}{(k+2)!} + \frac{(k+1)2^{k+1}}{(k+3)!}\) | A1 | Correct expression |
| \(= 1 - 2^{k+1}\left(\frac{1}{(k+2)!} - \frac{k+1}{(k+3)!}\right)\) | M1 | Use of part (a) |
| \(= 1 - 2^{k+1} \cdot \frac{2}{(k+3)!} = 1 - \frac{2^{k+2}}{(k+3)!}\) | A1 | Completion with conclusion statement |
## Question 6:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{1}{(k+2)!} - \frac{k+1}{(k+3)!} = \frac{k+3}{(k+3)!} - \frac{k+1}{(k+3)!}$ | M1 | Common denominator of $(k+3)!$ |
| $= \frac{(k+3)-(k+1)}{(k+3)!} = \frac{2}{(k+3)!}$ | A1 | Completion to given result |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| **Base case:** $n=1$: LHS $= \frac{1 \times 2}{3!} = \frac{2}{6} = \frac{1}{3}$; RHS $= 1 - \frac{4}{3!} = 1 - \frac{2}{3} = \frac{1}{3}$ ✓ | B1 | Verify $n=1$ |
| **Assume** $\sum_{r=1}^{k} \frac{r \times 2^r}{(r+2)!} = 1 - \frac{2^{k+1}}{(k+2)!}$ | M1 | Statement of assumption for $n=k$ |
| **For $n=k+1$:** Add $\frac{(k+1)2^{k+1}}{(k+3)!}$ to both sides | M1 | Attempt to add $(k+1)$th term |
| $= 1 - \frac{2^{k+1}}{(k+2)!} + \frac{(k+1)2^{k+1}}{(k+3)!}$ | A1 | Correct expression |
| $= 1 - 2^{k+1}\left(\frac{1}{(k+2)!} - \frac{k+1}{(k+3)!}\right)$ | M1 | Use of part (a) |
| $= 1 - 2^{k+1} \cdot \frac{2}{(k+3)!} = 1 - \frac{2^{k+2}}{(k+3)!}$ | A1 | Completion with conclusion statement |
---6 (a) Show that $\frac { 1 } { ( k + 2 ) ! } - \frac { k + 1 } { ( k + 3 ) ! } = \frac { 2 } { ( k + 3 ) ! }$.\\
(b) Prove by induction that, for all positive integers $n$,
$$\sum _ { r = 1 } ^ { n } \frac { r \times 2 ^ { r } } { ( r + 2 ) ! } = 1 - \frac { 2 ^ { n + 1 } } { ( n + 2 ) ! }$$
(6 marks)
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\hfill \mbox{\textit{AQA FP2 2010 Q6 [8]}}