FP2 June 2012 Q6

EdexcelOld spec11 marksSeries

(a) Express \(\dfrac{1}{r(r + 2)}\) in partial fractions. (2)

(b) Hence prove, by the method of differences, that \[\sum_{r=1}^{n}\frac{1}{r(r + 2)} = \frac{n(an + b)}{4(n + 1)(n + 2)}\] where \(a\) and \(b\) are constants to be found. (6)

Scheme	Marks
\(\dfrac{1}{r(r + 2)} = \dfrac{1}{2}\left(\dfrac{1}{r} - \dfrac{1}{r + 2}\right) = \dfrac{1}{2r},\ -\dfrac{1}{2r + 4}\)	B1,B1oe
	(2)

Mark scheme (b)

Scheme	Marks
\(r = 1:\ \ \dfrac{1}{2}\left(\dfrac{1}{1} - \dfrac{1}{3}\right)\) \(r = 2:\ \ \dfrac{1}{2}\left(\dfrac{1}{2} - \dfrac{1}{4}\right)\) \(r = 3:\ \ \dfrac{1}{2}\left(\dfrac{1}{3} - \dfrac{1}{5}\right)\) \(r = n - 1:\ \ \dfrac{1}{2}\left(\dfrac{1}{n - 1} - \dfrac{1}{n + 1}\right)\)	M1
\(r = n:\ \ \dfrac{1}{2}\left(\dfrac{1}{n} - \dfrac{1}{n + 2}\right)\)	A1
Summing: \(\displaystyle\sum_{r=1}^{n}\dfrac{1}{r(r + 2)} = \dfrac{1}{2}\left(1 + \dfrac{1}{2} - \dfrac{1}{n + 1} - \dfrac{1}{n + 2}\right)\)	M1 A1
\(= \dfrac{1}{2}\left(\dfrac{3(n + 1)(n + 2) - 2(n + 1) - 2(n + 2)}{2(n + 1)(n + 2)}\right) = \dfrac{n(3n + 5)}{4(n + 1)(n + 2)}\)	M1 A1cao
	(6)

Notes

1^st M1 must include at least 4 out of 5 of (\(r =\)) 1, 2, 3 and \(n - 1\), \(n\)

1^st A1 require all terms that do not cancel to be accurate

2^nd M1 Summed expression involving all terms that do not cancel

2^nd A1 Correct expression

3^rd M1 for attempt to find single fraction

Mark scheme (c)

Scheme	Marks
\(\displaystyle\sum_{r=1}^{2n}\dfrac{1}{r(r + 2)} = \dfrac{2n(6n + 5)}{4(2n + 1)(2n + 2)}\)	B1oe
\(S_{2n} - S_n = \dfrac{2n(6n + 5)}{4(2n + 1)(2n + 2)} - \dfrac{n(3n + 5)}{4(n + 1)(n + 2)}\) \(= \dfrac{n(6n + 5)(n + 2) - n(3n + 5)(2n + 1)}{4(n + 1)(n + 2)(2n + 1)}\)	M1
\(= \dfrac{n(6n^2 + 17n + 10 - 6n^2 - 13n - 5)}{4(n + 1)(n + 2)(2n + 1)} = \dfrac{n(4n + 5)}{4(n + 1)(n + 2)(2n + 1)}\) (ag)	A1 cso
	(3)
	(11 marks)

Notes

1^st M1 for expression for \(S_{2n} - S_n\)