| Exam Board | OCR MEI |
|---|---|
| Module | Further Pure Core (Further Pure Core) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove summation with fractions |
| Difficulty | Standard +0.8 This is a Further Maths proof by induction question requiring students to first determine constants through particular cases, then execute a non-trivial inductive proof involving factorial manipulation. The algebraic step in the induction requires careful handling of the general term (n/(n+1)!) and combining fractions with factorial denominators, which is more sophisticated than standard A-level induction proofs. The 9-mark allocation and Further Pure content place it moderately above average difficulty. |
| Spec | 4.01a Mathematical induction: construct proofs |
| Answer | Marks | Guidance |
|---|---|---|
| 11 | (i) | 1 b |
| Answer | Marks |
|---|---|
| Verify (or solve) a(cid:32)b(cid:32) 1 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1a |
| 2.2a | Two values of n(cid:116)2 to give |
| Answer | Marks | Guidance |
|---|---|---|
| 11 | (ii) | (cid:167)1 1(cid:183) |
| Answer | Marks |
|---|---|
| true for k + 1 hence true for n = 2, 3, 4, … | B1 |
| Answer | Marks |
|---|---|
| [7] | 1.1 |
| Answer | Marks |
|---|---|
| 2.5 | n |
Question 11:
11 | (i) | 1 b
E.g. n = 2 gives (cid:32)a–
2 2
5 b
and n = 3 gives (cid:32)a–
6 6
Verify (or solve) a(cid:32)b(cid:32) 1 | M1
A1
[2] | 3.1a
2.2a | Two values of n(cid:116)2 to give
two simultaneous equations
11 | (ii) | (cid:167)1 1(cid:183)
Result holds for n(cid:32) 2 (cid:168) (cid:32) (cid:184)
(cid:169)2 2(cid:185)
1 2 3 k(cid:16)1 1
Assume (cid:14) (cid:14) (cid:14)(cid:125)(cid:14) (cid:32)1(cid:16)
2! 3! 4! k! k!
1 2 3 k(cid:16)1 k
(cid:14) (cid:14) (cid:14)(cid:125)(cid:14) (cid:14)
2! 3! 4! k! (cid:11)k(cid:14)1(cid:12)!
1 k
(cid:32)1(cid:16) (cid:14)
k! (cid:11)k(cid:14)1(cid:12)!
e
1 k (cid:167)(cid:11)k(cid:14)1(cid:12)(cid:16)k (cid:183)
RHS is 1(cid:16) (cid:14) (cid:32)1(cid:16)(cid:168) p(cid:184)
k! (cid:11)k(cid:14)1(cid:12)! (cid:168) (cid:11)k(cid:14)1(cid:12)! (cid:184)
(cid:169) (cid:185)
(cid:167) 1 (cid:183) S
which is 1(cid:16)(cid:168) (cid:184)
(cid:168)(cid:11)k(cid:14)1(cid:12)! (cid:184)
(cid:169) (cid:185)
The result is true for n = 2. If true for n = k it is also
true for k + 1 hence true for n = 2, 3, 4, … | B1
E1
M1A1
c
M1
A1
E1
[7] | 1.1
2.1
m
2.1
1.1
i
1.1
2.2a
2.5 | n
e
Assume true for n = k
Add correct next term to both
sides
1 minus fraction with correct
denominator on RHS
Showing that this is the given
result with k + 1 replacing k
Complete argument\begin{enumerate}[label=(\roman*)]
\item It is conjectured that
$$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + ... + \frac{n-1}{n!} = a - \frac{b}{n!},$$
where $a$ and $b$ are constants, and $n$ is an integer such that $n \geq 2$.
By considering particular cases, show that if the conjecture is correct then $a = b = 1$. [2]
\item Use induction to prove that
$$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + ... + \frac{n-1}{n!} = 1 - \frac{1}{n!} \text{ for } n \geq 2.$$ [7]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Pure Core Q11 [9]}}