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```					Further Pure 1

Summation of finite Series
Sigma notation
 In the last lesson we met the following rules.
1)    1 + 2 + 3 + …… + n = (n/2)(n+1)
2)    12 + 22 + 32 + …… + n2 = (n/6)(n+1)(2n+1)
3)    13 + 23 + 33 + …… + n3 = (n2/4) (n+1)2
 We can write long summations like the ones above
using sigma notation.
n
1  2  3  .......  n   r
r 1
n
1  2  3  .......  n   r 2
2    2     2             2

r 1
n
1  2  3  .......  n   r 3
3    3     3             3

r 1
Sigma notation
 The r acts as a counter starting at 1 (or whatever is stated under
the sigma sign) and running till you get to n (on top of the sigma
sign).
 Each r value generates a term and then you simply add up all the
terms.
4

 2r  1  3  5  7  9  24
r 1

 The terms in the example above come from
r=1             2×1+1 = 3
r=2             2×2+1 = 5
r=3             2×3+1 = 7
r=4             2×4+1 = 9
 The 4 on top of the sigma sign tells us to stop when r = 4.
Questions
 Here are some questions for you to try and find the
values of.

5

 (r 2  4)  5  8  13  20  29  75
r 1
6

 5r  20  25  30  75
r 4
5

 (r
r 2
3
 r )  10  30  68  130  238
Sigma notation
 We can now remember the identities that we met
sigma notation.

n
n
1  2  3  .......  n   r  (n  1)
r 1 2
n
n
1  2  3  .......  n   r  (n  1)( 2n  1)
2    2    2            2        2

r 1 6
n
n2
13  23  33  .......  n3   r 3  (n  1)2
r 1   4
Using Nth terms
 Use the nth term to find the following summation.
6
n                   6
 r  6 (n  1)(2n  1)  6 (6  1)(2  6  1)  91
1
2

 The summation only works if you sum from 1 to n.
 How would you calculate the next example.
8       8      3
82         32

4
r 3   r 3   r 3  (8  1)2  (3  1)2  190
1       1      4          4
 Here the sum goes from r = 4, to r = 8.
 This means you do not want the terms for r = 1, 2 & 3.
 So the answer will be the sum to 8 minus the sum to 3.
Rules of summing series
 Here are 2 rules that you need to be familiar with.
 There is a numerical example followed by a general rule
 k and a represent random constants.
n

 5r  5  10  15  .......... ..... 5n         n          n
an
 ar  a r  2 (n  1)
r 1

 5(1  2  3  .......... .....  n)
n
5n                           r 1    r 1
 5 r        (n  1)
r 1     2
n

 7  7  7  7  .......... ..... 7                  n          n

 k  k  r  kn
r 1

 7(1  1  1  .......... .....  1)
r 1       r 1
n
 7 1  7n
r 1
Example
 These results can be used to find the sum to n of lots
of different series.

 First break the            n               n          n

summation up.          
r 1
(r  3r)   r  3 r
3

r 1
3

r 1

   Next use the general                   n2            3n
 (n  1)2       (n  1)
formula.                                4             2
   Here (n/4)(n+1) is a                   n
 (n  1)[n(n  1)  6]
factor                                 4
   Next just multiply out                 n
 (n  1)(n 2  n  6)
and collect up like                    4
terms.                                 n
 (n  1)(n  2)(n  3)
   Finally the expression                 4
will factorise.
Question
 Try this question
What is (1 2)  (2  3)  (3  4)  .......... .  (30  31)
30            30       30

 r(r  1)   r   r
r 1           1
2

1

n                   n
 (n  1)( 2n  1)  (n  1)
6                   2
30                          30
     (30  1)( 2  30  1)     (30  1)
6                           2
 5  31 61  15  31
 9920
Question
 Here the sum
What is (6  7)  (7  8)  (8  9)  .......... .  (30  31)
starts at r = 6.    30            30             5

 This is not as       r(r  1)   r(r  1)   r(r  1)
r 6          r 1          r 1
complicated as                            5 2 5 
it may seem.                     9920    r   r 
            
 1      1   
 All you need to                          n                  n        
do is take of the                9920   (n  1)( 2n  1)  (n  1) 
6                  2        
first 5 terms.                           5                     5        
 9920   (5  1)( 2  5  1)  (5  1) 
 So the sum                               6                     2        
from 6 to 30 is                          5            5     
the sum to 30                    9920    6  11   6 
6            2     
minus the sum                    9920  (55  15)
to 5.                            9850
Questions
 Here are some questions for you to find the nth
terms of.
 The solutions are on the next two slides
n
a)    (6r 2  2r  4)
r 1
n
b)    (4r
r 1
3
 3r  4)
n
c)    (r  1)(r  2)
r 1
n
d)    r(r  1)(r  2)
r 1
Solutions
n


n

 (6r          2
 2r  4)                     ( 4r 3  3r  4)
r 1                                           r 1
n               n           n
  4r  3r  4
n                n           n
  6r 2   2r   4
3

r 1              r 1        r 1             r 1             r 1        r 1
n                  n           n                n                n            n
 6  r  2 r  4  1
2
 4 r  3 r  4 1
3

r 1                 r 1        r 1              r 1            r 1          r 1

6n(n  1)(2n  1) 2n(n  1)                    4n            3n
                                4n              (n  1)2     (n  1)  4n
6                2                      4             2
 n(n  1)(2n  1)  n(n  1)  4n              2n(n  1)2  3n(n  1)  8n
 2n3  3n2  n  n2  n  4n                   n(2(n  1)2  3(n  1)  8)
 2n3  4n2  6n                                n(2n2  4n  2  3n  3  8)
 2n(n 2  2n  3)                              n(2n2  n  9)
Solutions
n                                       n

 (r  1)(r  2)
r 1
 r(r  1)(r  2)
r 1
n         n           n                 n         n               n
  r   3r   2
2
  r   3r   2r
3               2

r 1       r 1        r 1             r 1       r 1            r 1
n               n           n           n               n               n
  r  3  r  2 1
2
  r  3 r  2 r
3               2

r 1            r 1        r 1        r 1            r 1            r 1

n                   3n                      n2            3n                     2n
 (n  1)( 2n  1)     (n  1)  2n           (n  1)2      (n  1)( 2n  1)     (n  1)
6                    2                      4              6                      2
n                                           n
 [(n  1)( 2n  1)  9(n  1)  12]          (n  1)[n( n  1)  2(2n  1)  4]
6                                           4
n                                           n
 [2n2  3n  1  9n  9  12]                (n  1)[n 2  n  4n  2  4]
6                                           4
n                                           n
 [2n2  12n  22]                            (n  1)[n 2  5n  6]
6                                           4
n                                           n
 [n2  6n  11]                              (n  1)(n  2)(n  3)
3                                           4
Summation of a finite Series
 When Carl Friedrich Gauss was a boy in elementary
100 numbers.
   S100 = 1 + 2 + 3 + …………… + 100
   Gauss had a flash of mathematical genius and
realised that the sum had 50 pairs of 101
   Therefore S100 = 50 × 101
= 5 050
   From this we can come up with the formula for the
sum of the first n numbers.
   Sn = (n/2)(n+1)
   We have met this result a few times already.
Method of differences
 We can prove the same result using a
different method.
 The method of differences.
Example 1
 Use the method of differences to find the sum
to 30 of the following example.
1    1     1
i) Show that         
r r  1 r(r  1)
1      1       1                1
ii) Hence find                  ....... 
1 2 2  3 3  4              30  31

 Solution to part ii is on the next 2 slides.
 You covered adding fractions in C2 and
should be able to get the answer.
Example 1
 We can use the identity           1   1    1                  1      30
1        n
1  1 
          .......                                
to re-arrange the question. 1 2 2  3 3  4             30  31 r 1 r(r  1) r 1  r r  1
1 1
   Now write the summation out long hand.                           
1 2
Starting with r = 1.                                             
1 1
2 3
   Then r = 2,3 etc.                                                  1 1
 
   Write out the last 2 or 3 terms.                                   3 4
 ...  ...  ...
 Having written out the full summation you                                   1   1
    
can spot that parts of the sum cancel.                                     29 30
1   1
 The bits that are left do not cancel and we                                  
30 31
can sort out the sum.                                                     1
1 30

31 31

30
1      30
 r(r  1)  31
r 1
Example 2
 In this next example we will find the sum to n.
2   3     1        r4
i) Show that           
r r  1 r  2 r(r  1)(r  2)

n
r4
ii) Hence find 
r 1 r(r  1)(r  2)

 Solution to part ii is on the next 2 slides.
 You covered adding fractions in C2 and
should be able to get the answer.
Example 2
 We can use the identity
n
r4            n
2       3   1 
 r(r  1)(r  2)    r  r  1  r  2 
r 1                 
to re-arrange the question.          r 1

2 3 1
        Now write the summation out long hand.                   
1 2 3
Starting with r = 1.
2 3 1
  
        Then r = 2,3 etc.                                         2 3 4
        Write out the last 3 terms.                               2 3 1
  
3 4 5
        Having written out the full summation you
2 3 1
can spot that parts of the sum cancel.                   
4 5 6
        The bits that are left do not cancel and we             ...  ...  ...
can sort out the algebra.                                      2      3   1
           
n - 2 n 1 n
n
r4          2 3 2    1     3   1
 r(r  1)(r  2)
r 1
             
1 2 2 n 1 n 1 n  2                      
2     3
 
1
3   2    1
n -1 n n 1
                                           2     3     1
2 n 1 n  2                                        
n n 1 n  2

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