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Further Pure 1
Revision Topic 5: Sums of Series
The OCR syllabus says that candidates should:
Candidates should be able to:
(a) use the standard results for Σr , Σr2 , Σr3 to find related sums;
(b) use the method of differences to obtain the sum of a finite series;
(c) recognise, by direct consideration of the sum to n terms, when a series is convergent, and find
the sum to infinity in such cases.
Section 1: Using standard formulae for r, r 2 , r 3
You need to learn the formula:
n
1
r 2 n(n 1)
r 1
The following formulae are given in the formula book:
n
1
r 2 6 n(n 1)(2n 1)
r 1
n
1
r 3 4 n2 (n 1)2 .
r 1
n
Note: 1 n (this should be obvious to you – if it isn’t then you need to remember it!).
r 1
The above formulae can be used to find the sum of series.
Worked examination question (Edexcel June 2004)
n
1
(a) Show that (r 1)(r 5) =
r 1 6
n(n+ 7)(2n + 7). (4)
40
(b) Hence calculate the value of (r 1)( r 5)
r 10
(2)
Solution:
n n
a) (r 1)(r 5) (r 2 6r 5) (expanding out brackets)
r 1 r 1
We next split up the summation into several separate sums:
n n n n
(r 1)(r 5) r 2 6 r 51
r 1 r 1 r 1 r 1
So, using the results from above:
n
1 1 1
(r 1)(r 5) 6 n(n 1)(2n 1) 6 2 n(n 1) 5n = 6 n(n 1)(2n 1) 3n(n 1) 5n
r 1
1
We can try factorising this by taking out n as a factor:
6
n
1
(r 1)(r 5) 6 n (n 1)(2n 1) 18(n 1) 30
r 1
Expanding out the contents of the square brackets gives:
n
1 1
(r 1)(r 5) 6 n 2n2 3n 1 18n 18 30 6 n 2n2 21n 49
r 1
The contents of the squared brackets can now be factorised. We get the result:
n
1
(r 1)(r 5) = 6 n(n+ 7)(2n + 7).
r 1
40 40 9
b) (r 1)(r 5) (r 1)(r 5) (r 1)(r 5)
r 10 r 1 r 1
So:
40
1 1
(r 1)(r 5) 6 40(47)(87) 6 9 16 25
r 10
Therefore,
40
(r 1)(r 5) 27260 600 26660 .
r 10
Examination question: Edexcel June 2002
Prove that
n
6( r
r 1
2
1) (n 1)n(2n + 5).
Examination question (AQA June 2005)
a) Use standard formulae to show that
n
1
r 2 (r 1) 12n(n2 1)(3n 2) .
r 1
b) Use the result from part (a) to find the value of
11
r 2 (r 1) .
r 4
Examination question (AQA 2001)
a) Find the sum of the integers from 1 to 300 inclusive.
b) Evaluate:
300
(r 2 r )
r 1
Examination question (OCR May 2004)
(i) Use the formula for r 3 to show that
13 23 33 ... (2n 1)3 n2 (2n 1)2 .
(ii) Show also that
23 43 63 ... (2n 2)3 2n2 (n 1)2 .
(iii) Hence find the sum of the series
13 23 33 43 ... (2n 2)3 (2n 1)3 ,
simplifying your answer.
Section 2: Method of differences
Some series can be summed using a difference method.
Worked examination question (AQA 2006)
a) Show that
1 1 2r 1
2
r 2
(r 1) 2
r (r 1) 2
b) Hence find the sum of the first n terms of the series
3 5 7
2 2 2 2 ...
1 2
2 2
2 3 3 4
Solution:
a) Writing with a common denominator, we get
1 1 (r 1)2 r2
2 2
r 2 (r 1)2 r (r 1)2 r (r 1)2
Therefore,
1 1 r 2 2r 1 r 2 2r 1
2
r 2
(r 1) 2
r (r 1)
2 2
r (r 1)2
b) We wish to find
n
2r 1
r 2 (r 1)2 .
r 1
Using the result from (a), this is equivalent to finding
n
1 1
r 2 (r 1)2 .
r 1
Substituting r = 1, 2, 3, … into this summation gives:
1 1 1 1 1 1 1 1 1 1
2 2 2 2 2 2 ... 2 2
1 2 2 3 3 4 (n 1)
2
n n
(n 1)2
Therefore,
n
1 1 1
r 2 (r 1)2 = 1 (n 1)2 .
r 1
Note: The series given in the above question is convergent. As the value of n increases (i.e. as the
number of terms gets bigger and bigger), the sum of the series approaches 1 (since the value of
1
will tend to zero).
(n 1) 2
We write
1 1
r 2 (r 1)2 = 1.
r 1
Worked examination question (Edexcel Jan 2004)
(a) Show that (r + 1)3 – (r – 1)3 Ar2 + B, where A and B are constants to be found.
n
(b) Prove by the method of differences that r
r 1
2
= 1
6
n(n + 1)(2n + 1), n > 1.
Solution
a) On removing the brackets we find that
(r + 1)3 – (r – 1)3 = r 3 3r 2 3r 1 r 3 3r 2 3r 1 = 6r 2 2
So, A = 6 and B = 2.
n n
b) So, (6r 2 2) (r 1)3 (r 1)3 .
r 1 r 1
Working out the right hand side of this equation (by substituting r = 1, 2, 3, etc) gives:
n
(6r 2 2) ( 23 03 ) ( 33 13 ) ( 43 23 ) ( 53 33 )
r1 1st term 2nd term 3rd term 4th term
( 63 43 ) ... (n3 (n 2)3 ) ((n 1)3 (n 1)3 )
5th term n1th term nth term
i.e.
n
(6r 2 2) (n 1)3 n3 1
r 1
But the left-hand side of this equation is:
n n n n
(6r 2 2) 6 r 2 2 6 r 2 2n .
r 1 r 1 r 1 r 1
Therefore:
n
6 r 2 2n = (n 1)3 n3 1
r 1
n
So: 6 r 2 = (n 1)3 n3 1 2n (n3 3n2 3n 1) n3 1 2n
r 1
n
i.e. 6 r 2 = 2n3 3n2 n n(2n2 3n 1) n(2n 1)(n 1) .
r 1
So we get the result:
n
r
r 1
2
= 1 n(n + 1)(2n + 1).
6
Examination question (Adapted from Edexcel June 2005)
2 1 1
(a) Show that =
4 r 1 2r 1 2r 1
2
n
2 1
(b) Hence, or otherwise, prove that =1– .
r 1 4r 1
2
2n 1
20
2
(c) Hence find the exact value of
r 11 4r 1
2
.
Examination question (adapted from Edexcel Jan 2005)
1 1 1
(a) Show that .
r (r 2) 2r 2(r 2)
(b) Hence prove, by the method of differences, that
n
4 n(3n 5)
r (r 2) = (n 1)( n 2) .
r 1
100
4
(c) Find the value of
r 50 r (r 2)
, to 4 decimal places.
Examination question (AQA Jan 2005)
a) Show that
r 2 (r 1)2 (r 1)2 r 2 4r 3 .
100
b) Use the method of differences to find the value of r3 .
r 50
Examination question (AQA June 2004)
a) Show that
1 1 2
.
r (r 1) (r 1)(r 2) r (r 1)(r 2)
b) Hence find the sum of the series
1 1 1 1
... ,
1 2 3 2 3 4 3 4 5 30 31 32
giving your answer as a rational number.
Examination Question (OCR January 2005)
4
You are given that f (r ) .
(r 1)(r 3)
2 2
(i) Show that f (r ) .
r 1 r 3
n
(ii) Hence find f (r ) . (You need not express your answer as a single fraction.)
r 1
(iii) Show that the series in part (ii) is convergent, and state its sum to infinity.
Worked examination question (OCR May 2005)
n
1
a) Show that (3r 1)2 n(6n2 3n 1) .
r 1 2
1
b) (i) Given that f ( r ) , show that
r (r 1)
2
f (r 1) f (r ) .
r (r 1)(r 2)
n
1
(ii) Hence find r (r 1)(r 2) .
r 1
Solution:
n n n n n
a) (3r 1)2 (9r 2 6r 1) 9 r 2 6 r 1
r 1 r 1 r 1 r 1 r 1
Using the standard formulae given in section 1 of these revision notes, we see that
n
1 1
(3r 1)2 9 6 n(n 1)(2n 1) 6 2 n(n 1) n
r 1
3
= n(n 1)(2n 1) 3n( n 1) n .
2
Taking out ½n as a factor, we get:
n
1
(3r 1)2 2 n 3(n 1)(2n 1) 6(n 1) 2
r 1
1
2
n 6n 2 9n 3 6n 6 2
1
2
n 6n 2 3n 1
1 1
b) If f ( r ) , then f (r 1) .
r (r 1) (r 1)(r 2)
1 1
So, f (r 1) f (r ) = - .
(r 1)( r 2) r (r 1)
r r2 2
Therefore, f (r 1) f (r ) = .
r (r 1)(r 2) r (r 1)(r 2) r (r 1)(r 2)
n
1 1 n 2 1 n
(ii) r (r 1)(r 2) 2 r (r 1)(r 2) 2
= ( f (r 1) f (r )) .
r 1 r 1 r 1
Therefore,
n
1
1
r (r 1)(r 2) = 2 ( f (2) f (1)) ( f (3) f (2) ) ( f (4) f (3) ) ... ( f (n 1) f (n) )
r 1
n
1 1
i.e. r (r 1)(r 2) =
2
f (n 1) f (1) .
r 1
But,
1 1 1
f (1) and f (n 1) .
1 2 2 (n 1)(n 2)
Therefore
n
1 1 1 1 1 1 1
r (r 1)(r 2) =
2
f (1) f (n 1) 4 2( n 1)( n 2) .
2 2 (n 1)( n 2)
r 1
Examination question (AQA June 2003)
a) Show that
1 1 r
r ! (r 1)! (r 1)!
b) Hence find
n
r
(r 1)!
r 1
Hint: In part (a), use the result that (r + 1)! = (r + 1)r!
