1. Sequences & the Binomial Theorem Chapter:___

2. Sequences Section:____

3. Sequences
Sequence: a function whose domain is the set of positive integers.(A sequence is displayed as a list of ordered values called terms that follows a specific pattern: 𝒂 𝟏 , 𝒂 𝟐 , 𝒂 𝟑 , …, 𝒂 𝒏
general term: formula for 𝑛 𝑡ℎ term of sequence
Calculator functions & commands for sequences:
𝒏! 𝑛 factorial To calculate 5!= 1 ∙ 2 ∙ 3 ∙ 4 ∙ 5 =120
:seq( command To find 𝑛 terms Enter seq(formula,X,start,end)
:sum( command To find sum of 𝑛 terms. Enter sum(seq(formula,X,start,end)
1st term
2nd term
3rd term
𝑛𝑡ℎ term
TI-83/84: MATH → PRB
TI-83/84: 2nd STAT → OPS
TI-83/84: 2nd STAT → MATH

4. Sequences
Recursive sequence: a sequence in which the first term(or first few terms) is given and the 𝑛 𝑡ℎ term formula involves one or more preceding terms.
Notation: 𝑎 𝑛−1 means one term before 𝑎 𝑛 𝑎 𝑛+1 means one term after 𝑎 𝑛
Components of Summation notation: p. 643
to 𝑛 𝑡ℎ term.
𝑎 𝑘 = 𝑎 1 + 𝑎 2 + 𝑎 3 +⋯ 𝑎 𝑛
𝑘=1 𝑛
Add terms of sequence ( 𝑎 𝑘 )
from 1𝑠𝑡 term

5. Sequences
Example: Find the first 5 terms of the sequence whose general term is: 𝑎 𝑛 = −1 𝑛 (2𝑛)
First 5 terms:
Or seq(formula,X,start,end) seq( −1 X 2X ,X,1,5)
𝑎 1 =
− ∙1 =−2
𝑎 2 =
− ∙2 = 4
Do you notice the specific pattern of this sequence?
𝑎 3 =
− ∙3 =−6
𝑎 4 =
− ∙4 = 8
𝑎 5 =
− ∙5 =−10

6. Sequences
Example: Find the 𝑛 𝑡ℎ term of the sequence , 3 4 , 4 9 , 5 16 ,…
𝑎 1 = 2 1 , 𝑎 2 = 3 4 , 𝑎 3 = 4 9 , 𝑎 4 = 5 16 , …, 𝑎 𝑛 = ? ?
Numerator: 1+1, 2+1, 3+1, 4+1, ….
Denominator: 1∙1, 2∙2, 3∙3, 4∙4, ….
Formula for 𝑛 𝑡ℎ term:
𝑛+1
𝑛∙𝑛
𝑎 𝑛 = 𝑛+1 𝑛 2

7. Sequences
Example: Find the first 5 terms of the recursive sequence. 𝑎 1 =2, 𝑎 2 =6, 𝑎 𝑛 = 𝑎 𝑛−1 ∙ 𝑎 𝑛−2
𝑎 1 = 𝑎 2 =
𝑎 𝑛 = 𝑎 𝑛−1 ∙ 𝑎 𝑛−2 2
(1 term before 𝑎 𝑛 )∙(2 terms before 𝑎 𝑛 ) 6
𝑎 3 = 𝑎 2 ∙ 𝑎 1 =
6 2 = 12
𝑎 4 = 𝑎 3 ∙ 𝑎 2 =
12 6 = 72
𝑎 5 = 𝑎 4 ∙ 𝑎 3 =
=864

9. Arithmetic Sequences Section:____

10. Arithmetic Sequences 1
Arithmetic Sequence(AS): a sequence in which the difference 𝒅 between consecutive terms is always the same number.
Once you know 𝒂 𝟏 and 𝒅, you can find term 𝒂 𝟐 by adding 𝒅 to 𝒂 𝟏 and so on.
To find common difference 𝒅 for consecutive terms 𝒂 𝒏−𝟏 , 𝒂 𝒏 : 𝒅= 𝒂 𝒏 − 𝒂 𝒏−𝟏 nonconsecutive terms 𝒂 𝒌 , 𝒂 𝒏 : 𝒅= 𝒂 𝒏 − 𝒂 𝒌 𝒏−𝒌
To find 𝒏 𝒕𝒉 term of an AS : 𝒂 𝒏 = 𝒂 𝟏 + 𝒏−𝟏 𝒅
To find Sum of First 𝒏 terms of an AS: 𝑺 𝒏 = 𝒏 𝟐 [ 𝒂 𝟏 + 𝒂 𝒏 ] or 𝑺 𝒏 = 𝒏 𝟐 [𝟐 𝒂 𝟏 +(𝒏−𝟏)𝒅]
𝑎 1
𝑎 2
𝑎 3 …
𝑎 𝑛−1
𝑎 𝑛 +𝑑 +𝑑 +𝑑 +𝑑 +𝑑

11. Arithmetic Sequences 2
Example: Find the common difference 𝒅 and find first 4 terms of the Arithmetic Sequence 𝑎 𝑛 ={ 𝑛}
Find terms.
Find 𝒅.
On TI-83/84
(1)=
7 10
seq(( 1 2 )+( 1 5 )X, X, 1, 4)
𝑎 1 =
(2)=
9 10
𝑎 2 =
(3)=
11 10
𝑎 3 =
(4)=
13 10
𝑎 4 =
= 9 10 − 7 10 =
2 10 =
1 5
𝒅= 𝒂 𝒏 − 𝒂 𝒏−𝟏

12. Arithmetic Sequences 3
Example: Find 𝒂 𝟏 , 𝒅, and specified formulas for Arithmetic Sequence having: 𝑡ℎ term is 3; 𝑛𝑑 term is 71
Find 𝒅 (nonconsecutive terms):
Find 𝒂 𝟏 .
Find recursive formula:
Find nth term formula:
(71−3) (22−5) =
𝑎 𝑛 − 𝑎 𝑘 𝑛−𝑘 = 4
Solve 𝑎 5 = 𝒂 𝟏 + 𝑛−1 (𝒅) for 𝒂 𝟏 .
𝑎 5 − 𝑛−1 (𝒅)= 𝒂 𝟏
3− 5−1 4 =
−13
𝑎 𝑛 = 𝑎 𝑛−1 +𝒅=
𝑎 𝑛−1 +4
𝒂 𝒏 = 𝒂 𝟏 +(𝒏−1)𝒅
𝑎 𝑛 =−13+ 𝑛−1 4 =
4𝑛−17

13. Arithmetic Sequences 4 Example: Find the sum: 4+6+8+⋯+80 Find 𝒅:+𝟐 +𝟐
Example: Find the sum: ⋯+80
Find 𝒅:
Find number of terms 𝒏 in the sum:
Find sum 𝑺 𝒏 = 𝒏 2 [ 𝒂 𝟏 + 𝒂 𝒏 ]:
𝒂 𝟏
𝒂 𝒏
𝒅= 𝑎 𝑛 − 𝑎 𝑛−1
𝒅=6−4=2
Solve 𝒂 𝒏 = 𝒂 𝟏 + 𝒏−1 𝒅 for 𝒏.
80=4+(𝒏−1)(2)
76=(𝒏−𝟏)(2)
38=𝒏−𝟏
39=𝒏
𝑺 39 = =
1638

14. Arithmetic Sequences 5 Example: Find the sum:
𝑛=1
156
(−2𝑛+1)
Example: Find the sum:
Use 𝑺 𝒏 = 𝒏 2 2 𝒂 𝟏 + 𝒏−1 𝒅 to find sum.
Find the sum.
TI-83/84: Enter sum(seq(−2X+1, X, 1, 156)
Summation shows 𝒏=𝟏𝟓𝟔.
Find 𝒅 from 𝒂 𝟏 and 𝒂 𝟐 .
𝒂 𝟏 =−2 1 +1=−𝟏
𝒂 𝟐 =−2 2 +1=−𝟑
𝒅= 𝒂 𝟐 − 𝒂 𝟏
𝒅= −𝟑 − −𝟏 =−𝟐
𝑺 156 = 𝟏𝟓𝟔 2 2 −𝟏 + 𝟏𝟓𝟔−1 −𝟐 =
−24336

16. Geometric Sequences Section:____

17. Geometric Sequences 1
Geometric Sequence(GS): a sequence in which consecutive terms have the same common ratio 𝒓.
You can find term 𝒂 𝟐 by multiplying 𝒓 to 𝒂 𝟏 and so on.
To find common ratio 𝒓 for: consecutive terms 𝑎 𝑛 , 𝑎 𝑛− nonconsecutive terms 𝑎 𝑛 , 𝑎 𝑘
To find 𝒏 𝒕𝒉 term of a GS:
To find Sum of First 𝒏 terms of a GS:
Infinite Geometric Series: diverges if 𝒓 >𝟏 , therefore sum does not exist. (Sum gets larger as terms get larger.) converges if 𝒓 <𝟏 , therefore sum does exist. (Sum approaches limit as terms get smaller.)
𝑎 1
𝑎 2
𝑎 3 …
𝑎 𝑛−1
𝑎 𝑛 ∙𝑟 ∙𝑟 ∙𝑟 ∙𝑟 ∙𝑟
𝒓= 𝒂 𝒏 𝒂 𝒌
(𝒏−𝒌)
𝒓= (𝒂 𝒏 ) ( 𝒂 𝒏−𝟏 )
𝒂 𝒏 = 𝒂 𝟏 𝒓 𝒏−𝟏
𝑺 𝒏 = 𝒂 𝟏 ∙ (𝟏− 𝒓 𝒏 ) (𝟏−𝒓)
𝑺 ∞ = (𝒂 𝟏 ) (𝟏−𝒓)

18. Geometric Sequences 2
Example: Find the 7th term of the Geometric Sequence.
Find 𝒓.
Find 𝒂 𝟕 . Use 𝒂 𝒏 = 𝒂 𝟏 𝒓 𝒏−𝟏
∙(𝟑)
∙(𝟓)
1 2 , , , …
𝒓= 𝑎 𝑎 1 = = 3 5
𝒂 𝟕 = 𝟏 𝟐 𝟑 𝟓 (𝟕−1) =

19. Geometric Sequences 3
Example: Find the 𝑛𝑡ℎ term of the Geometric Sequence described.
The 𝑛 𝑡ℎ term is the formula of the sequence 𝒂 𝒏 = 𝒂 𝟏 (𝒓) 𝒏−1 You are given 𝒓. Need to find 𝒂 𝟏 .
Solve for 𝒂 𝟏 .
𝑎 7 =2916 , 𝑟=−3
𝑎 7 = 𝒂 𝟏 (−3) 𝟕−1
2916= 𝒂 𝟏 (−3) 𝟕−1
2916= 𝒂 𝟏 (729)
4= 𝒂 𝟏
𝒂 𝒏 = 𝑎 1 (𝑟) 𝑛−1 =
4 (−3) 𝑛−1

20. Geometric Sequences 4
Example: Find the sum of the Geometric Sequence. Round to three decimal places.
TI-83/84: Enter sum(seq(
𝑛=1 20 [(2)∙ 𝑛 ]
start
end
𝑛 𝑡ℎ term
X , X, 1, 20)
𝑺 𝒏 =
𝑺 𝒏 =
5.981

21. Geometric Sequences 5
Example: If the infinite geometric series converges, find the sum. …
Find ratio 𝒓. …
𝒓= ( 𝟏𝟓 𝟐 ) (𝟓) = 𝟑 𝟐 . Since ratio |𝒓|>𝟏, terms get larger, series diverges.
𝒓= ( 𝟏𝟓 𝟒 ) (𝟓) = 𝟑 𝟒 . Since ratio |𝒓|<𝟏, terms get smaller, series converges.
𝑺 ∞ = (𝒂 𝟏 ) (𝟏−𝒓)
= (𝟓) [𝟏− 𝟑 𝟒 ] = 20

23. Binomial Thereom Section:____

24. Binomial Theorem 1
Binomial Theorem: formula for expanding binomial expressions (𝒂+𝒃) 𝒏 with positive integer powers 𝒏. Number of terms is 𝒏+1.
To evaluate 𝒏 𝒓 : 𝒏 𝒓 = (𝒏!) 𝒓! [ 𝒏−𝒓 !]
TI-83/84:
To expand (𝒂+𝒃) 𝒏 will require a pattern based on the expansion formula: (𝒂+𝒃) 𝒏 = 𝑘=0 𝒏 𝒏 𝑘 (𝒂) 𝒏−𝑘 (𝒃) 𝑘 =
MATH → PRB →
𝒏 𝑪 𝒓
𝒏 0 (𝒂) 𝒏 (𝒃) 0 + 𝒏 1 (𝒂) 𝒏−1 (𝒃) 1 + 𝒏 2 (𝒂) 𝒏−2 (𝒃) 2 +⋯+ 𝒏 𝒏 (𝒂) 0 (𝒃) 𝒏
1𝑠𝑡 term 𝑘=0
2𝑛𝑑 term 𝑘=1
3𝑟𝑑 term 𝑘=2
(𝑛+1)𝑠𝑡 term 𝑘=𝑛

25. Binomial Theorem 2 Example: Evaluate the expression 10 8 .
Use 𝒏 𝒓 = (𝒏!) 𝒓! [ 𝒏−𝒓 !] .
TI-83/84: Enter
10 8 = (10!) 8! [ 10−8 !] =
(10!) 8! 2! 𝟏 𝟓 𝟏
= (1∙2∙3∙4∙5∙6∙7∙8∙9∙10) 1∙2∙3∙4∙5∙6∙7∙8 1∙2 = 45 10
MATH → PRB →
𝒏 𝑪 𝒓 8

26. Binomial Theorem 3 Example: Expand the expression ( 𝑥 2 +2 𝑦 3 ) 4 .
There are 4+1 terms having form 𝒏 𝑘 (𝒂) 𝒏−𝑘 (𝒃) 𝑘 : 1st term 2nd term 3rd term 4th term 5th term
= ( 𝑥 2 ) 𝟒 (2 𝑦 3 ) 𝟎 =
(1) (𝑥 8 )(1) =
𝑥 8
= ( 𝑥 2 ) 𝟑 (2 𝑦 3 ) 𝟏 =
4 𝑥 𝑦 =
8 𝑥 6 𝑦 3
= ( 𝑥 2 ) 𝟐 (2 𝑦 3 ) 𝟐 =
6 𝑥 𝑦 =
24 𝑥 4 𝑦 6
= ( 𝑥 2 ) 𝟏 (2 𝑦 3 ) 𝟑 =
4 𝑥 𝑦 =
32 𝑥 2 𝑦 9
= ( 𝑥 2 ) 𝟎 (2 𝑦 3 ) 𝟒 =
𝑦 12 =
16 𝑦 12
𝑥 8 +8 𝑥 6 𝑦 𝑥 4 𝑦 𝑥 2 𝑦 𝑦 12