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ksgo2016
@ksgo2016
July 2022
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Ребята, помогите
Начиная с какого номера члены геометрической прогрессии
1) 32,16,8, ... меньше 0,01
2) 1/3 , 2/3, 4/3, ... больше 50?
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Начиная с какого номера члены геометрической прогрессии
1) 32,16,8, ... меньше 0,01
b1=32 q=1/2 bn=b1
·q^(n-1) b1·q^(n-1)<1/100 32·(1/2 )^(n-1)<1/100
2^(5-n+1)<1/100
6-n<log ₂(1/100) n>6-log ₂(1/100) n>6+log ₂(100) n>6+2log ₂(10)
3<log ₂(10)<4 (2³=8; 2⁴=16) n>6+2(3) n>12
2) 1/3 , 2/3, 4/3, ... больше 50?
b1=1/3 q=2 bn=b1q^(n-1)>50 (1/3)
·2^(n-1)>50 2^(n-1)>150
n-1>log₂150 n>1+log₂150 7<log₂150 <8 ⇒n>1+7
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Answers & Comments
Verified answer
Начиная с какого номера члены геометрической прогрессии1) 32,16,8, ... меньше 0,01
b1=32 q=1/2 bn=b1·q^(n-1) b1·q^(n-1)<1/100 32·(1/2 )^(n-1)<1/100
2^(5-n+1)<1/100
6-n<log ₂(1/100) n>6-log ₂(1/100) n>6+log ₂(100) n>6+2log ₂(10)
3<log ₂(10)<4 (2³=8; 2⁴=16) n>6+2(3) n>12
2) 1/3 , 2/3, 4/3, ... больше 50?
b1=1/3 q=2 bn=b1q^(n-1)>50 (1/3)·2^(n-1)>50 2^(n-1)>150
n-1>log₂150 n>1+log₂150 7<log₂150 <8 ⇒n>1+7