# revision series

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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  51
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 ,
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 ) 
r1                    1st term          2nd term           3rd term           4th term

( 63  43 )  ...  (n3  (n  2)3 )  ((n  1)3  (n  1)3 )
5th term                  n1th 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
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                r2             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!

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