AS June 2023 Paper 1 Q7

AQACurrent spec7 marksSeries

(a) Show that, for all integers \(r\),\[\frac{1}{2r - 1} - \frac{1}{2r + 1} = \frac{2}{(2r - 1)(2r + 1)}\] [1 mark]

(b) Hence, using the method of differences, show that\[\sum_{r=1}^{n} \frac{1}{(2r - 1)(2r + 1)} = \frac{an}{bn + c}\]

where \(a\), \(b\) and \(c\) are integers to be determined. [4 marks]

(c) Hence, or otherwise, evaluate\[\frac{1}{1 \times 3} + \frac{1}{3 \times 5} + \frac{1}{5 \times 7} + \ \ldots\ + \frac{1}{99 \times 101}\] [2 marks]

Mark scheme (a)

Scheme	Marks	AO
Completes a rigorous argument to prove the required result. Must include the LHS, at least one intermediate step, and the RHS.	B1	2.1
	(1)

Typical solution

\[\begin{aligned}\frac{1}{2r - 1} - \frac{1}{2r + 1} &= \frac{2r + 1 - (2r - 1)}{(2r - 1)(2r + 1)} \\ &= \frac{2}{(2r - 1)(2r + 1)}\end{aligned}\]

Mark scheme (b)

Scheme	Marks	AO
Writes at least two pairs of subtracting fractions. Condone any consistent multiple of the correct fractions.	M1	1.1a
Writes at least one pair of cancelling fractions. Condone any consistent multiple of the correct fractions.	M1	1.1a
Correctly reduces the required expression to two terms. PI Condone any consistent multiple of the correct fractions.	A1	1.1b
Completes a reasoned argument using the method of differences to reach the required result. This mark is only available if at least the first two pairs of fractions and the last pair are shown. Accept an unsimplified fraction eg \(\dfrac{2n}{4n + 2}\)	R1	2.1
	(4)

Typical solution

\[\begin{aligned} &\sum_{r=1}^{n} \frac{2}{(2r - 1)(2r + 1)} \\ &= \sum_{r=1}^{n}\left(\frac{1}{2r - 1} - \frac{1}{2r + 1}\right) \\ &= \frac{1}{1} - \frac{1}{3} \\ &\quad + \frac{1}{3} - \frac{1}{5} \\ &\quad + \ldots\ldots\ldots \\ &\quad + \frac{1}{2n - 3} - \frac{1}{2n - 1} \\ &\quad + \frac{1}{2n - 1} - \frac{1}{2n + 1} \\ &= \frac{1}{1} - \frac{1}{2n + 1} \\ &= \frac{2n + 1 - 1}{2n + 1} \\ &= \frac{2n}{2n + 1}\end{aligned}\]\[\text{So} \quad \sum_{r=1}^{n} \frac{1}{(2r - 1)(2r + 1)} = \frac{n}{2n + 1}\]

Mark scheme (c)

Scheme	Marks	AO
Uses \(n = 50\) Condone 49 or 51	M1	3.1a
Obtains the correct exact value. oe eg \(0.\dot{4}95\dot{0}\) FT their \(\dfrac{50a}{50b + c}\)	A1F	1.1b
	(2)
	(7 marks)

Typical solution

\[\begin{aligned} &\frac{1}{1 \times 3} + \frac{1}{3 \times 5} + \frac{1}{5 \times 7} + \ldots + \frac{1}{99 \times 101} \\ &= \sum_{r=1}^{50} \frac{1}{(2r - 1)(2r + 1)} \\ &= \frac{50}{101}\end{aligned}\]