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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)
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4/8/16 Chapter 9 Review Algebra II
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Learning Objective To remember everything in Chapter 9!
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Arithmetic Sequence & Series
SEQUENCE/Explicit nth term formula π π = π 1 + πβ1 π Sum/SERIES Formula π π = π 2 ( π 1 + π π ) Summation Notation π=1 # πΈπ₯ππππππ‘ πΉππππ’ππ
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Geometric Sequence & Series
SEQUENCE/Explicit nth term formula π π = π 1 (π) πβ1 Sum/SERIES Formula (Finite) π π = π 1 (1β π π ) 1βπ Sum/SERIES Formula (Infinite) π= π 1 1βπ π <1 Converge π β₯1 Diverge
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Summation Notation Arithmetic Geometric π=1 # πΈπ₯ππππππ‘ πΉππππ’ππ
π=1 # π 1 (π) πβ1 π=1 # π 1 + πβ1 π
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Write the explicit formula & find the 30th term
1. β15, β10, β5, β¦ π π = π 1 + πβ1 π π π =β15+ πβ1 (5) π π =β15+5πβ5 π π =5πβ20 π 30 =5 30 β20 π 30 =130
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Find the missing terms 2. The sequence is arithmetic 17, ___, ___, ___, 41 + 6 + 6 + 6 + 6 23 29 35 π 1 = 17 n = 5 d = ___ π π = 41 π π = π 1 + πβ1 π 41=17+ 5β1 (π) 24=4π π=6
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Find the sum 3. n = 50 π 1 =3 1 +4=7 π π = π 50 =3 50 +4=154
π=1 50 (3π+4) 3. n = 50 π 1 =3 1 +4=7 π π = π 50 = =154 π 50 = 50 2 (7+(154)) π π = π 2 ( π 1 + π π ) =4025
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Solve 4. *** Number of terms = Top β Bottom + 1 n = 200 β 30 + 1 = 171
π= (4π+12) 4. *** Number of terms = Top β Bottom + 1 n = 200 β = 171 π π = π 2 ( π 1 + π π ) First term = 4(30) + 12 First term = 132 Last term = 4(200) + 12 Last term = 812 π 171 = ( ) π 171 =80712
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Write the Summation Notation & Find the Sum
β¦ + 112 n = ___ π 1 =16 π π =112 d = 6 Use π π = π 1 + πβ1 π to find n 112 = 16 + (n β 1)(6) 112 = n - 6 112 = 6n + 10 102 = 6n n = 17 π=1 17 (6π+10) π 17 =1088 π 17 = 17 2 (16+112) π π = π 2 ( π 1 + π π )
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Write the Summation Notation & Find the Sum
(-25)+(-12)+β¦+573 n = ___ π 1 =β38 π π =573 d = 13 Use π π = π 1 + πβ1 π to find n 573 = (n β 1)(13) 573 = n - 13 573 = 13n -51 624 = 13n n = 48 π=1 48 (13πβ51) π 48 =12840 π 48 = 48 2 (β38+573) π π = π 2 ( π 1 + π π )
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Write explicit formula & find 8th term
Geo: r = β 1 2 7. -6, 3, β 3 2 ,β¦ β π 1 π π = π 1 (π) πβ1 π π =β6 β πβ1 NOT π π = 3 πβ1 π 8 =β6 β β1 π 8 = 3 64
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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 π = π 1 (π) πβ1 64=4 (π) 5β1 16= (π) 4 π=Β±2 π 1 = 4 n = 5 r = ___ π π = 64
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Find the sum Infinite Geometric Series 9. 2β1+ 1 2 β 1 4 +β¦ = β1 2
9. 2β β 1 4 +β¦ r = π 2 π 1 = β1 2 π= π 1 1βπ π= 2 1β(β1/2) π <1 Converges π= 2 3/2 = 4 3 π=1 β 2 β πβ1
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Find the sum 10. 3β9+27+β¦ Infinite Geometric Series r = π 2 π 1 = β9 3
10. 3β9+27+β¦ Infinite Geometric Series r = π 2 π 1 = β9 3 π=β3 π β₯1 Diverges No Sum
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Find the sum π π = π 1 (1β π π ) 1βπ 11. π 8 = β3(1β (3) 8 ) 1β(3)
π=1 8 β3 (3) πβ1 π π = π 1 (1β π π ) 1βπ 11. π 8 = β3(1β (3) 8 ) 1β(3) n = 8 π 1 =β3 r = 3 π 8 = β3(1β(6561)) β2 π 8 = β3(β6560) β2 π 8 =β9840
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Arithmetic Sequence 12. π 1 =89 π 7 =71 What is π 20 ?
π π =89+ πβ1 (β3) π π = π 1 + πβ1 π π 20 =89+(20β1)(β3) π 7 = π 1 + 7β1 π π 20 =32 71 = d d = -3
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Geometric Sequence 13. π 1 =3 π 10 =59049 What is π 3 ? r = 3
π π = π 1 (π) πβ1 π π =3 (3) πβ1 π 10 = π 1 (π) 10β1 π 3 =3 (3) 3β1 59049=3 (π) 9 π 3 =3 (3) 2 19683= (π) 9 π 3 =27
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Expand 14. ( π¦ 2 β3π₯) 4 ( π¦ 2 ) 4 (β3π₯) 0 + ( π¦ 2 ) 3 (β3π₯) 1 4 +
14. ( π¦ 2 β3π₯) 4 ( π¦ 2 ) 4 (β3π₯) 0 1 + ( π¦ 2 ) 3 (β3π₯) 1 4 + ( π¦ 2 ) 2 6 (β3π₯) 2 + ( π¦ 2 ) 1 (β3π₯) 3 + ( π¦ 2 ) 0 (β3π₯) 4 4 1 π¦ 8 β 12 π¦ 6 π₯ + 54 π¦ 4 π₯ 2 β108 π¦ 2 π₯ 3 + 81 π₯ 4
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Find the 4th term 15. (3π₯β2π¦) 9 (3π₯) 9 (β2π¦) 0 + (3π₯) 8 9 (β2π¦) 1 +
15. (3π₯β2π¦) 9 (3π₯) 9 1 (β2π¦) 0 + (3π₯) 8 9 (β2π¦) 1 + 36 (3π₯) 7 (β2π¦) 2 + 84 (3π₯) 6 (β2π¦) 3 β π₯ 6 π¦ 3
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Assignment: W-S Review 2 Study for Test (Mon) Virtual Lab (Sun 6pm)
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