| Exam Board | AQA |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Finding n for given sum value |
| Difficulty | Standard +0.8 Part (a) requires partial fraction decomposition of a three-factor denominator and telescoping series manipulation—standard Further Maths techniques but with multiple algebraic steps. Part (b) is straightforward substitution and solving an inequality once (a) is complete. This is a typical Further Maths summation question requiring solid technique but no novel insight, placing it moderately above average difficulty. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| 13(a) | Uses partial fractions. | 3.1a |
| Answer | Marks | Guidance |
|---|---|---|
| OE | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| sum | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| versa) | 2.5 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| required result. AG | 2.1 | R1 |
| Subtotal | 5 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 13(b) | Obtains an inequality in n | |
| PI | 3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| PI | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| of n | 2.2a | A1 |
| Subtotal | 3 | |
| Question total | 8 | |
| Q | Marking instructions | AO |
Question 13:
--- 13(a) ---
13(a) | Uses partial fractions. | 3.1a | M1 | 1 A B C
≡ + +
(r –1)r(r+1) r –1 r r+1
1≡ Ar(r+1)+B(r –1)(r+1)+C(r –1)r
r =0: 1= –B⇒ B= –1
r =1: 1=2A⇒ A= 1
2
r = –1: 1=2C ⇒C = 1
2
n 1 n 1 1 1
∑ =∑ – +
(r–1)r(r+1) 2(r –1) r 2(r+1)
r=2 r=2
1 1 1
= – +
2(1) 2 2(3)
1 1 1
+ – +
2(2) 3 2(4)
1 1 1
+ – +
2(3) 4 2(5)
+......
1 1 1
+ – +
2(n–3) n–2 2(n–1)
1 1 1
+ – +
2(n–2) n–1 2n
1 1 1
+ – +
2(n–1) n 2(n+1)
n 1 1 1 1 1 1 1
∑ = − + + – +
(r –1)r(r+1) 2 2 4 2n n 2(n+1)
r=2
1 1 1
= – +
4 2n 2(n+1)
Obtains
1 1 1
– +
2(r –1) r 2(r+1)
OE | 1.1b | A1
Writes at least 3
consecutive rows of the
sum | 1.1a | M1
Uses the method of
differences showing at
least the first three and
last two terms (or vice
versa) | 2.5 | M1
Completes fully correct
working to reach the
required result. AG | 2.1 | R1
Subtotal | 5
Q | Marking instructions | AO | Marks | Typical solution
--- 13(b) ---
13(b) | Obtains an inequality in n
PI | 3.1a | M1 | ∑ n 1 > 0.24999
(r–1)r(r+1)
r=2
1 1 1
– + > 0.24999
4 2n 2(n+1)
1 1
0.00001 > –
2n 2(n+1)
1
0.00001 >
2n(n+1)
2n(n + 1) > 105
2n2 + 2n – 105 > 0
Solutions to 2n2 + 2n – 105 = 0 are
223.1 and –224.1
n > 223.1
n = 224
Rearranges their inequality
or equation to obtain a
quadratic inequality or
equation.
PI | 1.1a | M1
Deduces the correct value
of n | 2.2a | A1
Subtotal | 3
Question total | 8
Q | Marking instructions | AO | Marks | Typical solution\begin{enumerate}[label=(\alph*)]
\item Use the method of differences to show that
$$\sum_{r=2}^{n} \frac{1}{(r - 1)r(r + 1)} = \frac{1}{4} - \frac{1}{2n} + \frac{1}{2(n + 1)}$$
[5 marks]
\item Find the smallest integer $n$ such that
$$\sum_{r=2}^{n} \frac{1}{(r - 1)r(r + 1)} > 0.24999$$
[3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 2 2024 Q13 [8]}}