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

1r

33333

n

1r

22222

n

1r

rn.......321

rn.......321

rn.......321

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.

The terms in the example above come fromr = 1 2×1+1 = 3r = 2 2×2+1 = 5r = 3 2×3+1 = 7r = 4 2×4+1 = 9

The 4 on top of the sigma sign tells us to stop when r = 4.

2497531r24

1r

Questions Here are some questions for you to try and find the

values of.

238130683010 )rr(

75302520 r5

7529201385 )4r(

5

2r

3

6

4r

5

1r

2

Sigma notation We can now remember the identities that we met

last lesson and have mentioned already adding the sigma notation.

22n

1r

33333

n

1r

22222

n

1r

)1n(4

nrn.......321

)1n2)(1n(6

nrn.......321

)1n(2

nrn.......321

Using Nth terms

Use the nth term to find the following summation.

The summation only works if you sum from 1 to n. How would you calculate the next example.

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.

190)13(4

3)18(

4

8rrr

91)162)(16(6

6)1n2)(1n(

6

nr

22

223

1

38

1

38

4

3

6

1

2

Rules of summing series

)1n(2

n5r5

n)...............325(1

n5...............15105r5

n

1r

n

1r

n717

1)...............117(1

7...............7777

n

1r

n

1r

)1n(2

anra ar

n

1r

n

1r

knrk kn

1r

n

1r

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.

Example

3))(n21)(n(n4

n

)6n1)(n(n4

n

6]1)1)[n(n(n4

n

1)(n2

3n1)(n

4

n

r3r3r)(r

2

22

n

1r

n

1r

3n

1r

3

These results can be used to find the sum to n of lots of different series.

First break the summation up.

Next use the general formula.

Here (n/4)(n+1) is a factor Next just multiply out and

collect up like terms. Finally the expression will

factorise.

Question

9920

311561315

)130(2

30)1302)(130(

6

30

)1n(2

n)1n2)(1n(

6

n

rr)1r(r

)13(30...........4)(33)(22)(1 is What30

1

30

1

230

1r

Try this question

Question

9850

)1555(9920

62

5116

6

5 9920

)15(2

5)152)(15(

6

5 9920

)1n(2

n)1n2)(1n(

6

n 9920

rr 9920

)1r(r)1r(r)1r(r

)13(30...........9)(88)(77)(6 is What

5

1

5

1

2

5

1r

30

1r

30

6r

Here the sum starts at r = 6.

This is not as complicated as it may seem.

All you need to do is take of the first 5 terms.

So the sum from 6 to 30 is the sum to 30 minus the sum to 5.

Questions

)2r)(1r(r )d

)2r)(1r( )c

)4r3r4( )b

)4r2r6( )a

n

1r

n

1r

n

1r

3

n

1r

2

Here are some questions for you to find the nth terms of.

The solutions are on the next two slides

Solutions

3)2n2n(n

6n4n2n

4nnnn3n2n

4n1)n(n1)1)(2nn(n

4n2

1)2n(n

6

1)1)(2n6n(n

14r2r6

4r2r6

)4r2r6(

2

23

223

n

1r

n

1r

n

1r

2

n

1r

n

1r

n

1r

2

n

1r

2

)9nn2(n

)83n324nn2(n

)81)(n31)(n2(n

n81)(nn31)(nn2

n41)(n2

3n1)(n

4

4n

14r3r4

4r3r4

)4r3r4(

2

2

2

2

2

n

1r

n

1r

n

1r

3

n

1r

n

1r

n

1r

3

n

1r

3

Solutions

]11n6n[3

n

]22n12n[26

n

]129n91n3n[26

n

]12)1n(9)1n2)(1n[(6

n

n2)1n(2

n3)1n2)(1n(

6

n

12r3r

2r3r

)2r)(1r(

2

2

2

n

1r

n

1r

n

1r

2

n

1r

n

1r

n

1r

2

n

1r

3)2)(n1)(n(n4

n

]6n51)[n(n4

n

]42n4n1)[n(n4

n

]4)1n2(2)1n1)[n((n4

n

)1n(2

n2)1n2)(1n(

6

n3)1n(

4

n

r2r3r

r2r3r

)2r)(1r(r

2

2

22

n

1r

n

1r

2n

1r

3

n

1r

n

1r

2n

1r

3

n

1r

Summation of a finite Series

When Carl Friedrich Gauss was a boy in elementary school his teacher asked his class to add up the first 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.

Use the method of differences to find the sum to 30 of the following example.

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

3130

1.......

43

1

32

1

21

1 find Hence )ii

)1r(r

1

1r

1

r

1 that Show )i

Example 1

31

30

31

1 1

31

1

30

1

30

1

29

1

......... 4

1

3

1

3

1

2

1

2

1

1

1

1r

1

r

1

1)r(r

1

3130

1.......

43

1

32

1

21

1 n

1r

30

1r

31

30

1)r(r

130

1r

We can use the identity to re-arrange the question.

Now write the summation out long hand. Starting with r = 1.

Then r = 2,3 etc. Write out the last 2 or 3 terms. Having written out the full summation you

can spot that parts of the sum cancel. The bits that are left do not cancel and we

can sort out the sum.

Example 2

2))(r1r(r

4r find Hence )ii

2))(r1r(r

4r

2r

1

1r

3

r

2 that Show )i

n

1r

Solution to part ii is on the next 2 slides. You covered adding fractions in C2 and

should be able to get the answer.

In this next example we will find the sum to n.

Example 2

2n

1

1n

3

n

2

1n

1

n

3

1-n

2

n

1

1n

3

2-n

2

......... 6

1

5

3

4

2

5

1

4

3

3

2

4

1

3

3

2

2

3

1

2

3

1

2

2r

1

1r

3

r

2

2)1)(rr(r

4r n

1r

n

1r

2n

1

1n

2

2

3

2n

1

1n

3

1n

1

2

2

2

3

1

2

2)1)(rr(r

4rn

1r

We can use the identity to re-arrange the question.

Now write the summation out long hand. Starting with r = 1.

Then r = 2,3 etc. Write out the last 3 terms. Having written out the full summation you

can spot that parts of the sum cancel. The bits that are left do not cancel and we

can sort out the algebra.