FP2 June 2014 Q1 – Westie's Workshop

FP2 June 2014 Q1

EdexcelOld spec6 marksSeries

1.

(a) Express \(\dfrac{2}{(r + 2)(r + 4)}\) in partial fractions. (1)

(b) Hence show that \[\sum_{r=1}^{n}\frac{2}{(r + 2)(r + 4)} = \frac{n(7n + 25)}{12(n + 3)(n + 4)}\] (5)

Mark scheme (b)

Scheme	Marks
\(\displaystyle\sum_{r=1}^{n}\dfrac{2}{(r + 2)(r + 4)} = \sum_{r=1}^{n}\left(\dfrac{1}{r + 2} - \dfrac{1}{r + 4}\right)\)
\(= \dfrac{1}{3} - \dfrac{1}{5} + \dfrac{1}{4} - \dfrac{1}{6} + \ldots\ldots\) \(+ \dfrac{1}{n + 1} - \dfrac{1}{n + 3} + \dfrac{1}{n + 2} - \dfrac{1}{n + 4}\) Attempts at least the first 2 terms and at least the last 2 terms as shown. (May be implied by later work) Must start at 1 and end at \(n\)	M1
\(= \dfrac{1}{3} + \dfrac{1}{4} - \dfrac{1}{n + 3} - \dfrac{1}{n + 4}\) M1: Identifies their four fractions that do not cancel. If all terms are positive this mark is lost. A1: Correct four fractions	M1A1
\(= \dfrac{7}{12} - \dfrac{1}{n + 3} - \dfrac{1}{n + 4}\)
\(= \dfrac{7(n + 3)(n + 4) - 12(n + 4) - 12(n + 3)}{12(n + 3)(n + 4)}\) \(= \dfrac{7n^2 + 49n + 84 - 12n - 48 - 12n - 36}{12(n + 3)(n + 4)}\) Attempt to combine at least 3 fractions, 2 of which have a function of \(n\) in the denominator and expands the numerator. As a minimim, the product of 2 linear factors must be expanded in the numerator.	M1
\(= \dfrac{n(7n + 25)}{12(n + 3)(n + 4)}\) * cso Must be factorised. If worked with \(r\) instead of \(n\) throughout, deduct last mark only.	A1
	(5)
	(6 marks)

(b) Way 2

Scheme	Marks
\(\dfrac{7}{12} - \left(\dfrac{1}{n + 3} + \dfrac{1}{n + 4}\right)\)
\(= \dfrac{7}{12} - \left(\dfrac{n + 4 + n + 3}{(n + 3)(n + 4)}\right)\)
\(= \dfrac{7(n + 3)(n + 4) - 24n - 84}{12(n + 3)(n + 4)}\)
\(= \dfrac{7n^2 + 49n + 84 - 24n - 84}{12(n + 3)(n + 4)}\) Attempt to combine at least 3 fractions, 2 of which have a function of \(n\) in the denominator and expands the numerator. Min as above	M1
\(= \dfrac{n(7n + 25)}{12(n + 3)(n + 4)}\) * cso	A1

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