Standard +0.8 This is a Further Maths question requiring partial fractions with method of differences. Part (a) is straightforward partial fractions, but part (b) requires recognizing the telescoping pattern and careful bookkeeping of terms from r=4 to r=97. The technique is standard for Further Maths students, but the execution requires precision and multiple steps, placing it moderately above average difficulty.
\begin{enumerate}[label=(\alph*)]
\item Given that $$\frac{2}{(r + 1)(r + 2)(r + 3)} \equiv \frac{A}{(r + 1)(r + 2)} + \frac{B}{(r + 2)(r + 3)}$$
find the values of the integers \(A\) and \(B\)
[2 marks]
\item Use the method of differences to show clearly that $$\sum_{r=4}^{97} \frac{1}{(r + 1)(r + 2)(r + 3)} = \frac{89}{19800}$$
[4 marks]
Question 3:
--- 3(a) ---
3(a) | Forms identity using the
numerators from each side | AO1.1a | M1 | 2 A B
≡ +
(r+1)(r+2)(r+3) (r+1)(r+2) (r+2)(r+3)
⇒2≡A(r+3)+B(r+1)
⇒A=1 ,B=−1
Obtains the correct values of A
and B | AO1.1b | A1
(b) | Uses ‘their’ result from part (a) to
write fraction as sum of differences.
1
Ignore at this stage.
2 | AO1.1a | M1 | ∑ 97 1 = 1∑ 97 1 − 1
r=9 (r+1)(r+2)(r+3) 2r=9 (r+1)(r+2) (r+2)(r+3)
97 1 1
∑ −
(r+1)(r+2) (r+2)(r+3)
r=9
1
=
10×11
1 1
− +
11×12 11×12
1 1
− +
12×13 12×13
...
1 1
− +
98×99 98×99
1
−
99×100
1 1
= −
10×11 99×100
89
=
9900
97 1
∴∑
(r+1)(r+2)(r+3)
r=9
1 89
= ×
2 9900
89
=
19800
Clearly shows step of cancelling of
terms | AO2.4 | M1
Obtains correct two term difference
Ft ‘their’ values for A and B
provided that ‘their’ A = − ‘their’ B | AO1.1b | A1
States that method of differences
gives
97 1
2 × ∑
(r+1)(r+2)(r+3)
r=9
so divides their answer by 2 to
obtain correct rational solution from
fully correct working AG
(If student merely divides by 2
without justification withhold this
mark) | AO2.1 | R1
Total | 6
Q | Marking Instructions | AO | Marks | Typical Solution
\begin{enumerate}[label=(\alph*)]
\item Given that $$\frac{2}{(r + 1)(r + 2)(r + 3)} \equiv \frac{A}{(r + 1)(r + 2)} + \frac{B}{(r + 2)(r + 3)}$$
find the values of the integers $A$ and $B$
[2 marks]
\item Use the method of differences to show clearly that $$\sum_{r=4}^{97} \frac{1}{(r + 1)(r + 2)(r + 3)} = \frac{89}{19800}$$
[4 marks]
</end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 1 Q3 [6]}}