Sequences & the Binomial Theorem Chapter:___

Sequences Section:____

Sequences 1 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

Sequences 2 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

Sequences 3 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 1 2∙1 =−2 𝑎 2 = −1 2 2∙2 = 4 Do you notice the specific pattern of this sequence? 𝑎 3 = −1 3 2∙3 =−6 𝑎 4 = −1 4 2∙4 = 8 𝑎 5 = −1 5 2∙5 =−10

Sequences 4 Example: Find the 𝑛 𝑡ℎ term of the sequence. 2 1 , 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

Sequences 5 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 = 72 12 =864

Arithmetic Sequences Section:____

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

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

Arithmetic Sequences 3 Example: Find 𝒂 𝟏 , 𝒅, and specified formulas for Arithmetic Sequence having: 5 𝑡ℎ term is 3; 22 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

Arithmetic Sequences 4 Example: Find the sum: 4+6+8+⋯+80 Find 𝒅: +𝟐 +𝟐 Example: Find the sum: 4+6+8+⋯+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 = 39 2 4 + 80 = 1638

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

Geometric Sequences Section:____

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−1 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)

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 , 3 10 , 9 50 , … 𝒓= 𝑎 2 𝑎 1 = 3 10 1 2 = 3 5 𝒂 𝟕 = 𝟏 𝟐 𝟑 𝟓 (𝟕−1) = 729 31250

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

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)∙ 3 4 𝑛 ] start end 𝑛 𝑡ℎ term 2 3 4 X , X, 1, 20) 𝑺 𝒏 =5.980972728 𝑺 𝒏 = 5.981

Geometric Sequences 5 Example: If the infinite geometric series converges, find the sum. 5+ 15 2 + 45 4 +… Find ratio 𝒓. 5+ 15 4 + 45 16 +… 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

Binomial Thereom Section:____

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

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

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 = 4 0 ( 𝑥 2 ) 𝟒 (2 𝑦 3 ) 𝟎 = (1) (𝑥 8 )(1) = 𝑥 8 = 4 1 ( 𝑥 2 ) 𝟑 (2 𝑦 3 ) 𝟏 = 4 𝑥 6 2 𝑦 3 = 8 𝑥 6 𝑦 3 = 4 2 ( 𝑥 2 ) 𝟐 (2 𝑦 3 ) 𝟐 = 6 𝑥 4 4 𝑦 6 = 24 𝑥 4 𝑦 6 = 4 3 ( 𝑥 2 ) 𝟏 (2 𝑦 3 ) 𝟑 = 4 𝑥 2 8 𝑦 9 = 32 𝑥 2 𝑦 9 = 4 4 ( 𝑥 2 ) 𝟎 (2 𝑦 3 ) 𝟒 = 1 1 16 𝑦 12 = 16 𝑦 12 𝑥 8 +8 𝑥 6 𝑦 3 +24 𝑥 4 𝑦 6 +32 𝑥 2 𝑦 9 +16 𝑦 12