cho A =1/2+(1/2)^2+(1/2)^3+(1/2)^4+ …+(1/2)^98+(1/2)^99 Hãy so sánh A và 1 28/08/2021 Bởi Camila cho A =1/2+(1/2)^2+(1/2)^3+(1/2)^4+ …+(1/2)^98+(1/2)^99 Hãy so sánh A và 1
Ta có: A= 1/2+ ( 1/2) ^ 2+ …+ ( 1/2) ^ 99 = 1/2+ 1/ 2^ 2+ 1/2^ 3+…+ 1/2^99 2. A= 1+ 1/2+ 1/2^ 2+…+ 1/2^98 2.A- A= ( 1+ 1/2+ 1/2^ 2+…+ 1/2^98) – ( 1/2+ 1/ 2^ 2+ 1/2^ 3+…+ 1/2^99) A = 1 – 1/2^99 < 1 ( Vì 1/2^99> 0) Chúc học tốt! Bình luận
`A = 1/2 + (1/2) ^2 + (1/2)^3 + (1/2)^4 +….. + (1/2)^{98} + (1/2)^{99}` `2A = 1 + 1/2 + 1/4 + 1/8 + …. + 1/{2^{97}} + 1/{2^{98}}` `2A – A = (1 + 1/2 + 1/4 + 1/8 + …. + 1/{2^{97}} + 1/{2^{98}})-[1/2 + (1/2) ^2 + (1/2)^3 + (1/2)^4 +….. + (1/2)^{98} + (1/2)^{99}]` `A = 1 – 1/{2^{99}}` $⇒$ $A < 1$ vì `1/{2^{99}} > 0` Bình luận
Ta có:
A= 1/2+ ( 1/2) ^ 2+ …+ ( 1/2) ^ 99
= 1/2+ 1/ 2^ 2+ 1/2^ 3+…+ 1/2^99
2. A= 1+ 1/2+ 1/2^ 2+…+ 1/2^98
2.A- A= ( 1+ 1/2+ 1/2^ 2+…+ 1/2^98) – ( 1/2+ 1/ 2^ 2+ 1/2^ 3+…+ 1/2^99)
A = 1 – 1/2^99 < 1 ( Vì 1/2^99> 0)
Chúc học tốt!
`A = 1/2 + (1/2) ^2 + (1/2)^3 + (1/2)^4 +….. + (1/2)^{98} + (1/2)^{99}`
`2A = 1 + 1/2 + 1/4 + 1/8 + …. + 1/{2^{97}} + 1/{2^{98}}`
`2A – A = (1 + 1/2 + 1/4 + 1/8 + …. + 1/{2^{97}} + 1/{2^{98}})-[1/2 + (1/2) ^2 + (1/2)^3 + (1/2)^4 +….. + (1/2)^{98} + (1/2)^{99}]`
`A = 1 – 1/{2^{99}}`
$⇒$ $A < 1$ vì `1/{2^{99}} > 0`