FP2 – Chapter 2 – Method of Differences Exam Questions
FP2 June 2009
1.
(a) Express
1
in partial fractions.
r (r  2)
(1)
n
(b) Hence show that
4
 r (r  2)
=
r 1
n(3n  5)
.
(n  1)( n  2)
(5)
FP2 June 2010
1.
(a) Express
3
in partial fractions.
(3r  1)(3r  2)
(2)
(b) Using your answer to part (a) and the method of differences, show that
n
3
 (3r  1)(3r  2)
r 1
=
3n
.
2(3n  2)
(3)
1000
(c) Evaluate

r  100
3
, giving 
your answer to 3 significant figures.
(3r  1)(3r  2)
(2)
FP2 June 2011
4.
Given that
(2r + 1)3 = Ar3 + Br2 + Cr + 1,
(a) find the values of the constants A, B and C.
(2)
(b) Show that
(2r + 1)3 – (2r – 1)3 = 24r2 + 2.
(2)
(c) Using the result in part (b) and the method of differences, show that
n
r
r 1
2
=
1
n(n + 1)(2n + 1).
6
(5)
FP2 June 2012
6.
(a) Express
1
in partial fractions.
r (r  2)
(2)
(b) Hence prove, by the method of differences, that
n
1
 r (r  2)
r 1
=
n(an  b)
,
4(n  1)( n  2)
where a and b are constants to be found.
(6)
(c) Hence show that
2n

r  n 1
1
n(4n  5)
=
.
r (r  2) 4(n  1)( n  2)( 2n  1)
(3)
n
24. Prove by the method of differences that
r
r 1
2
=
1
6
n(n + 1)(2n + 1), n > 1.
(6)
[P4 January 2004 Qn 1]
Solutions
Question 1
Question 2
Question 3
Question 4