1. Summation of finite Series
Further Pure 1
Summation of finite Series

2. Sigma notation In the last lesson we met the following rules.
1) …… + n = (n/2)(n+1)
2) …… + n2 = (n/6)(n+1)(2n+1)
3) …… + n3 = (n2/4) (n+1)2
We can write long summations like the ones above using sigma notation.

3. 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 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.

4. Questions
Here are some questions for you to try and find the values of.

5. Sigma notation
We can now remember the identities that we met last lesson and have mentioned already adding the sigma notation.

6. 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.

7. 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.

8. Example
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.

9. Question
Try this question

10. Question 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.

11. Questions Here are some questions for you to find the nth terms of.
The solutions are on the next two slides

12. Solutions

13. Solutions

14. 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 = …………… + 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.

15. Method of differences
We can prove the same result using a different method.
The method of differences.

16. Example 1
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.

17. Example 1 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.

18. Example 2 In this next example we will find the sum to n.
Solution to part ii is on the next 2 slides.
You covered adding fractions in C2 and should be able to get the answer.

19. Example 2 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.