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Homework Log Fri. 4/8 Lesson Rev Learning Objective:

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1 Homework Log Fri. 4/8 Lesson Rev Learning Objective:
To remember everything in Chapter 9! Hw: W-S Review 2 Study for Test (Mon) Virtual Lab (Sun 6pm)

2 4/8/16 Chapter 9 Review Algebra II

3 Learning Objective To remember everything in Chapter 9!

4 Arithmetic Sequence & Series
SEQUENCE/Explicit nth term formula π‘Ž 𝑛 = π‘Ž 1 + π‘›βˆ’1 𝑑 Sum/SERIES Formula 𝑆 𝑛 = 𝑛 2 ( π‘Ž 1 + π‘Ž 𝑛 ) Summation Notation 𝑛=1 # 𝐸π‘₯𝑝𝑙𝑖𝑐𝑖𝑑 πΉπ‘œπ‘Ÿπ‘šπ‘’π‘™π‘Ž

5 Geometric Sequence & Series
SEQUENCE/Explicit nth term formula π‘Ž 𝑛 = π‘Ž 1 (π‘Ÿ) π‘›βˆ’1 Sum/SERIES Formula (Finite) 𝑆 𝑛 = π‘Ž 1 (1βˆ’ π‘Ÿ 𝑛 ) 1βˆ’π‘Ÿ Sum/SERIES Formula (Infinite) 𝑆= π‘Ž 1 1βˆ’π‘Ÿ π‘Ÿ <1 Converge π‘Ÿ β‰₯1 Diverge

6 Summation Notation Arithmetic Geometric 𝑛=1 # 𝐸π‘₯𝑝𝑙𝑖𝑐𝑖𝑑 πΉπ‘œπ‘Ÿπ‘šπ‘’π‘™π‘Ž
𝑛=1 # π‘Ž 1 (π‘Ÿ) π‘›βˆ’1 𝑛=1 # π‘Ž 1 + π‘›βˆ’1 𝑑

7 Write the explicit formula & find the 30th term
1. βˆ’15, βˆ’10, βˆ’5, … π‘Ž 𝑛 = π‘Ž 1 + π‘›βˆ’1 𝑑 π‘Ž 𝑛 =βˆ’15+ π‘›βˆ’1 (5) π‘Ž 𝑛 =βˆ’15+5π‘›βˆ’5 π‘Ž 𝑛 =5π‘›βˆ’20 π‘Ž 30 =5 30 βˆ’20 π‘Ž 30 =130

8 Find the missing terms 2. The sequence is arithmetic 17, ___, ___, ___, 41 + 6 + 6 + 6 + 6 23 29 35 π‘Ž 1 = 17 n = 5 d = ___ π‘Ž 𝑛 = 41 π‘Ž 𝑛 = π‘Ž 1 + π‘›βˆ’1 𝑑 41=17+ 5βˆ’1 (𝑑) 24=4𝑑 𝑑=6

9 Find the sum 3. n = 50 π‘Ž 1 =3 1 +4=7 π‘Ž 𝑛 = π‘Ž 50 =3 50 +4=154
𝑛=1 50 (3𝑛+4) 3. n = 50 π‘Ž 1 =3 1 +4=7 π‘Ž 𝑛 = π‘Ž 50 = =154 𝑆 50 = 50 2 (7+(154)) 𝑆 𝑛 = 𝑛 2 ( π‘Ž 1 + π‘Ž 𝑛 ) =4025

10 Solve 4. *** Number of terms = Top – Bottom + 1 n = 200 – 30 + 1 = 171
𝑛= (4𝑛+12) 4. *** Number of terms = Top – Bottom + 1 n = 200 – = 171 𝑆 𝑛 = 𝑛 2 ( π‘Ž 1 + π‘Ž 𝑛 ) First term = 4(30) + 12 First term = 132 Last term = 4(200) + 12 Last term = 812 𝑆 171 = ( ) 𝑆 171 =80712

11 Write the Summation Notation & Find the Sum
… + 112 n = ___ π‘Ž 1 =16 π‘Ž 𝑛 =112 d = 6 Use π‘Ž 𝑛 = π‘Ž 1 + π‘›βˆ’1 𝑑 to find n 112 = 16 + (n – 1)(6) 112 = n - 6 112 = 6n + 10 102 = 6n n = 17 𝑛=1 17 (6𝑛+10) 𝑆 17 =1088 𝑆 17 = 17 2 (16+112) 𝑆 𝑛 = 𝑛 2 ( π‘Ž 1 + π‘Ž 𝑛 )

12 Write the Summation Notation & Find the Sum
(-25)+(-12)+…+573 n = ___ π‘Ž 1 =βˆ’38 π‘Ž 𝑛 =573 d = 13 Use π‘Ž 𝑛 = π‘Ž 1 + π‘›βˆ’1 𝑑 to find n 573 = (n – 1)(13) 573 = n - 13 573 = 13n -51 624 = 13n n = 48 𝑛=1 48 (13π‘›βˆ’51) 𝑆 48 =12840 𝑆 48 = 48 2 (βˆ’38+573) 𝑆 𝑛 = 𝑛 2 ( π‘Ž 1 + π‘Ž 𝑛 )

13 Write explicit formula & find 8th term
Geo: r = βˆ’ 1 2 7. -6, 3, βˆ’ 3 2 ,… ↑ π‘Ž 1 π‘Ž 𝑛 = π‘Ž 1 (π‘Ÿ) π‘›βˆ’1 π‘Ž 𝑛 =βˆ’6 βˆ’ π‘›βˆ’1 NOT π‘Ž 𝑛 = 3 π‘›βˆ’1 π‘Ž 8 =βˆ’6 βˆ’ βˆ’1 π‘Ž 8 = 3 64

14 Find the missing terms 8. The sequence is geometric 4, ___, ___, ___, 64 x -2 x -2 x -2 x -2 -8 16 -32 8 16 32 x 2 x 2 x 2 x 2 π‘Ž 𝑛 = π‘Ž 1 (π‘Ÿ) π‘›βˆ’1 64=4 (π‘Ÿ) 5βˆ’1 16= (π‘Ÿ) 4 π‘Ÿ=Β±2 π‘Ž 1 = 4 n = 5 r = ___ π‘Ž 𝑛 = 64

15 Find the sum Infinite Geometric Series 9. 2βˆ’1+ 1 2 βˆ’ 1 4 +… = βˆ’1 2
9. 2βˆ’ βˆ’ 1 4 +… r = π‘Ž 2 π‘Ž 1 = βˆ’1 2 𝑆= π‘Ž 1 1βˆ’π‘Ÿ 𝑆= 2 1βˆ’(βˆ’1/2) π‘Ÿ <1 Converges 𝑆= 2 3/2 = 4 3 𝑛=1 ∞ 2 βˆ’ π‘›βˆ’1

16 Find the sum 10. 3βˆ’9+27+… Infinite Geometric Series r = π‘Ž 2 π‘Ž 1 = βˆ’9 3
10. 3βˆ’9+27+… Infinite Geometric Series r = π‘Ž 2 π‘Ž 1 = βˆ’9 3 π‘Ÿ=βˆ’3 π‘Ÿ β‰₯1 Diverges No Sum

17 Find the sum 𝑆 𝑛 = π‘Ž 1 (1βˆ’ π‘Ÿ 𝑛 ) 1βˆ’π‘Ÿ 11. 𝑆 8 = βˆ’3(1βˆ’ (3) 8 ) 1βˆ’(3)
𝑛=1 8 βˆ’3 (3) π‘›βˆ’1 𝑆 𝑛 = π‘Ž 1 (1βˆ’ π‘Ÿ 𝑛 ) 1βˆ’π‘Ÿ 11. 𝑆 8 = βˆ’3(1βˆ’ (3) 8 ) 1βˆ’(3) n = 8 π‘Ž 1 =βˆ’3 r = 3 𝑆 8 = βˆ’3(1βˆ’(6561)) βˆ’2 𝑆 8 = βˆ’3(βˆ’6560) βˆ’2 𝑆 8 =βˆ’9840

18 Arithmetic Sequence 12. π‘Ž 1 =89 π‘Ž 7 =71 What is π‘Ž 20 ?
π‘Ž 𝑛 =89+ π‘›βˆ’1 (βˆ’3) π‘Ž 𝑛 = π‘Ž 1 + π‘›βˆ’1 𝑑 π‘Ž 20 =89+(20βˆ’1)(βˆ’3) π‘Ž 7 = π‘Ž 1 + 7βˆ’1 𝑑 π‘Ž 20 =32 71 = d d = -3

19 Geometric Sequence 13. π‘Ž 1 =3 π‘Ž 10 =59049 What is π‘Ž 3 ? r = 3
π‘Ž 𝑛 = π‘Ž 1 (π‘Ÿ) π‘›βˆ’1 π‘Ž 𝑛 =3 (3) π‘›βˆ’1 π‘Ž 10 = π‘Ž 1 (π‘Ÿ) 10βˆ’1 π‘Ž 3 =3 (3) 3βˆ’1 59049=3 (π‘Ÿ) 9 π‘Ž 3 =3 (3) 2 19683= (π‘Ÿ) 9 π‘Ž 3 =27

20 Expand 14. ( 𝑦 2 βˆ’3π‘₯) 4 ( 𝑦 2 ) 4 (βˆ’3π‘₯) 0 + ( 𝑦 2 ) 3 (βˆ’3π‘₯) 1 4 +
14. ( 𝑦 2 βˆ’3π‘₯) 4 ( 𝑦 2 ) 4 (βˆ’3π‘₯) 0 1 + ( 𝑦 2 ) 3 (βˆ’3π‘₯) 1 4 + ( 𝑦 2 ) 2 6 (βˆ’3π‘₯) 2 + ( 𝑦 2 ) 1 (βˆ’3π‘₯) 3 + ( 𝑦 2 ) 0 (βˆ’3π‘₯) 4 4 1 𝑦 8 βˆ’ 12 𝑦 6 π‘₯ + 54 𝑦 4 π‘₯ 2 βˆ’108 𝑦 2 π‘₯ 3 + 81 π‘₯ 4

21 Find the 4th term 15. (3π‘₯βˆ’2𝑦) 9 (3π‘₯) 9 (βˆ’2𝑦) 0 + (3π‘₯) 8 9 (βˆ’2𝑦) 1 +
15. (3π‘₯βˆ’2𝑦) 9 (3π‘₯) 9 1 (βˆ’2𝑦) 0 + (3π‘₯) 8 9 (βˆ’2𝑦) 1 + 36 (3π‘₯) 7 (βˆ’2𝑦) 2 + 84 (3π‘₯) 6 (βˆ’2𝑦) 3 βˆ’ π‘₯ 6 𝑦 3

22 Assignment: W-S Review 2 Study for Test (Mon) Virtual Lab (Sun 6pm)


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