1. FM Series

2. 𝑥 2 +5𝑥+6 2𝑥(𝑥+3)+7(𝑥+3) 4 (𝑥+3) 2 −5(𝑥+3) Series KUS objectives
BAT Understand and write series using sigma notation
BAT Generate sequences
BAT Solve arithmetic series problems involving sigma notation
Starter factorise:
𝑥 2 +5𝑥+6
2𝑥(𝑥+3)+7(𝑥+3)
4 (𝑥+3) 2 −5(𝑥+3)

3. 1 𝑛 (2𝑟−1) The sum of odd numbers is
Notes on notation
The sum of odd numbers is
Sn = …. + (2n – 1)
We could write this as:
number of terms, the last term will be 2(n)-1
1 𝑛 (2𝑟−1)
r is the position of each term in the series
1 is the first position number, so this series starts at 2(1) -1 = 1
 is sigma in the ancient Greek alphabet and stands for ‘the sum of’

4. (Series are not always arithmetic)
WB1 Try these:
(Series are not always arithmetic)
1 𝑛 (3𝑟+5) =
…. + (3n+5)
1 10 (3𝑟+5) =
…. + 35
8 20 (3𝑟+5) =
…. + 65
1 6 (3 𝑟 2 +1) =
1 20 (4𝑟−1)
… =
𝑟 2
… =
1 25 (83−7𝑟)
… =
1 7 2 𝑟 3 =
All these series start at the first term

5. For an arithmetic series the ‘sum of terms’ is worked out
Notes on Sn
For an arithmetic series the ‘sum of terms’ is worked out
using this formula 𝑆𝑛 = 𝑛 2 2𝑎+ 𝑛−1 𝑑
Where: a is the first term, d is the common difference
and n is the number of terms. We assume the
sequence starts at the first term
If an arithmetic series is the sum of the same constant term each time
Then 𝑆𝑛 =𝑛𝑎 or 𝑛 𝑎 =𝑛𝑎

6. WB2 a) Evaluate 1 22 (3𝑟+5) using the formula for the sum of an arithmetic series
1 22 (3𝑟+5) =
…. + (3n+5) …
This is an arithmetic series,
using the formula for sum of terms
𝑆 𝑛 = n 2 2𝑎+ 𝑛−1 𝑑
a = 8
d = 3
n = 22
1 22 (3𝑟+5) = ×3
= 869

7. This is an arithmetic series, using the formula for sum of terms
WB2 b) Evaluate (7−2𝑟) using the formula for the sum of an arithmetic series. Explain why the answer is negative
1 40 (7−2𝑟) =
(-1) + (-3) + …. + (7-2(40))
This is an arithmetic series,
using the formula for sum of terms
𝑆 𝑛 = n 2 2𝑎+ 𝑛−1 𝑑
1 40 (7−2𝑟) = ×(−2)
a = 5
d = -2
n = 40
=−1360
This negative because most of the terms added up are negative

8. This is NOT an arithmetic series,
WB2 c) Evaluate 𝑟(𝑟−6) using the formula for the sum of an arithmetic series
This is NOT an arithmetic series,
You cannot use the formula for sum of terms
1 6 𝑟(𝑟−6) =
1(1-6) + 2(2-6) + 3(3-6) +4(4-6) + 5(5-6) + 0
= (-5) + (-8) + (-9) + (-8) + (-5)+0
= - 35

9. 1 𝑛 (3𝑟+4) =3 1 𝑛 𝑟 +4𝑛 Show that: 7 + 10 + 13 + 16 + 19 + …. + (3n+4)
WB3 key result
1 𝑛 (3𝑟+4) =3 1 𝑛 𝑟 +4𝑛
Show that:
1 𝑛 (3𝑟+4) =
…. + (3n+4)
Split into sums
= (3x1+ 4) + (3x2+ 4) + (3x3+ 4) + (3x4+ 4)+ … + (3n+ 4)
Split further
= (3x1+ 3x2 + 3x3 + 3x4 +… + 3n) + ( … + 4)
Split further still
= 3 ( … + n) + ( … + 4)
= 𝑛 𝑟 𝑛 𝑄𝐸𝐷

10. We will come back to this later
WB3 (cont)
1 𝑛 4 =4𝑛
Note that:
1 𝑛 (𝑎𝑟+𝑏) =𝑎 1 𝑛 𝑟 +𝑏𝑛
and:
We will come back to this later

11. Notes three key results
Use Sn = ½n[2a+(n-1)d] to find a formula in terms of n for the sum of the first n natural numbers
1 𝑛 𝑟 = 𝑛 2 2×1+(𝑛−1)×1 = 𝑛 2 𝑛+1
The formula for the sum of the first n squares is
1 𝑛 𝑟 2 = 𝑛 6 𝑛+1 (2𝑛+1)
The formula for the sum of the first n cubes is
1 𝑛 𝑟 3 = 𝑛 (𝑛+1) 2
These are in the exam booklet and proofs are not required

12. Try these: ½ (50)(51) = 1275 1275 1/6 (20)(21)(41) = ¼ (14)2 (15)2 =
1 𝑛 𝑟 = 𝑛 2 𝑛+1
1 𝑛 𝑟 2 = 𝑛 6 𝑛+1 (2𝑛+1)
1 𝑛 𝑟 3 = 𝑛 (𝑛+1) 2
WB4
Try these:
1 50 𝑟 =
½ (50)(51) =
1275
1 20 𝑟 2 =
1275
1/6 (20)(21)(41) =
1 14 𝑟 3 =
¼ (14)2 (15)2 =
1275

13. Remember: and: Calculate: WB5a 1 𝑛 4 =4𝑛 1 𝑛 (𝑎𝑟+𝑏) =𝑎 1 𝑛 𝑟 +𝑏𝑛
1 𝑛 𝑟 = 𝑛 2 𝑛+1
1 𝑛 𝑟 2 = 𝑛 6 𝑛+1 (2𝑛+1)
1 𝑛 𝑟 3 = 𝑛 (𝑛+1) 2
WB5a
1 𝑛 4 =4𝑛
Remember:
1 𝑛 (𝑎𝑟+𝑏) =𝑎 1 𝑛 𝑟 +𝑏𝑛
and:
Calculate:
1 46 (7𝑟−3) =
𝑟 −3(46)
=7[ 46 2 (46+1)]−3(46)

14. Calculate 1 25 𝑟= 3 1 25 𝑟 + 25(1) =3 (25)(26) 2 +50 =1025 1 25 (3𝑟+1)
1 𝑛 𝑟 = 𝑛 2 𝑛+1
1 𝑛 𝑟 2 = 𝑛 6 𝑛+1 (2𝑛+1)
1 𝑛 𝑟 3 = 𝑛 (𝑛+1) 2
WB5b
1 25 (3𝑟+1)
Calculate
1 25 𝑟= 𝑟 (1)
=3 (25)(26)
=1025

15. Calculate 1 28 30−5𝑟 1 28 30−5𝑟 =30×28−5 1 28 𝑟 =30×28−5 28 2 29 =434
1 𝑛 𝑟 = 𝑛 2 𝑛+1
1 𝑛 𝑟 2 = 𝑛 6 𝑛+1 (2𝑛+1)
1 𝑛 𝑟 3 = 𝑛 (𝑛+1) 2
WB 5c
Calculate −5𝑟
−5𝑟 =30×28− 𝑟
=30×28−
=434

16. WB6ab Find a) 50 100 𝑟 b) 10 25 𝑟 3 c) 5 30 ( 𝑟 2 −2) d) 11 25 𝑟(𝑟+1)
1 𝑛 𝑟 = 𝑛 2 𝑛+1
1 𝑛 𝑟 2 = 𝑛 6 𝑛+1 (2𝑛+1)
1 𝑛 𝑟 3 = 𝑛 (𝑛+1) 2
WB6ab Find a) 𝑟 b) 𝑟 c) ( 𝑟 2 −2) d) 𝑟(𝑟+1)
A neat trick is:
𝑎) 𝑟 = 𝑟 − 1 49 𝑟
= − = 3825
𝑏) 𝑟 3 = 𝑟 3 − 1 9 𝑟 3
= (26) 2 − =xxx

17. WB6c Find a) 50 100 𝑟 b) 10 25 𝑟 3 c) 5 30 ( 𝑟 2 −2) d) 11 25 𝑟(𝑟+1)
1 𝑛 𝑟 = 𝑛 2 𝑛+1
1 𝑛 𝑟 2 = 𝑛 6 𝑛+1 (2𝑛+1)
1 𝑛 𝑟 3 = 𝑛 (𝑛+1) 2
WB6c Find a) 𝑟 b) 𝑟 c) ( 𝑟 2 −2) d) 𝑟(𝑟+1)
𝑐) 𝑟 2 −2 = 𝑟 2 −2 − 𝑟 2 −2
= −2×30 − −2×4
= xxx

18. WB6d Find a) 50 100 𝑟 b) 10 25 𝑟 3 c) 5 30 ( 𝑟 2 −2) d) 11 25 𝑟(𝑟+1)
1 𝑛 𝑟 = 𝑛 2 𝑛+1
1 𝑛 𝑟 2 = 𝑛 6 𝑛+1 (2𝑛+1)
1 𝑛 𝑟 3 = 𝑛 (𝑛+1) 2
WB6d Find a) 𝑟 b) 𝑟 c) ( 𝑟 2 −2) d) 𝑟(𝑟+1)
𝑑) 𝑟(𝑟+1) = 𝑟 2 +𝑟 − 𝑟 2 +𝑟
= −
= − = xxx

19. WB7ab Try these:
1 10 (𝑟+1)(𝑟−2)
=245:
1 20 𝑟 2 (𝑟−1)
= 41230

20. Crucial points
1. Check your results When you find a sum of the first n terms of a series, it is a good idea to substitute n = 1, and perhaps n = 2 as well, to check your result.
2. Look for common factors When using standard results, there can be quite a lot of algebra involved in simplifying the result. Make sure you take out any common factors first, as this makes the algebra a lot simpler.
3. Be careful when summing a constant term
remember
1 𝑛 𝑘 =𝑘+𝑘+𝑘 𝑘=𝑘𝑛

21. KUS objectives
BAT Understand and write series using sigma notation
BAT Generate sequences
BAT Solve arithmetic series problems involving sigma notation
self-assess
One thing learned is –
One thing to improve is –

22. END