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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 7: cho A = 1/2 + (1/2)^2 + … + (1/2)^99 CM A < 1

Toán Lớp 7: cho A = 1/2 + (1/2)^2 + … + (1/2)^99
CM A < 1

Comments ( 2 )

  1. ~ gửi bạn ~
    Giải đáp:
    đ.p.c.m 
    Lời giải và giải thích chi tiết:
    A = 1/2 + (1/2)^2 + … + (1/2)^99 
    A = 1/2 + 1/2^2 + … + 1/2^99
    => 2A = 1 + 1/2 + 1/2^2 + … + 1/2^98
    => 2A – A = (1 + 1/2 + 1/2^2 + … + 1/2^98) – (1/2 + 1/2^2 + … + 1/2^99)
    => A = 1 – 1/2^99
    => A < 1

  2. Giải đáp:
    $A<1$
    Lời giải và giải thích chi tiết:
    $A=\dfrac12+\left(\dfrac12\right)^{\large2}\!\!\ +\,.\!.\!.+\ \left(\dfrac12\right)^{\large99}\\\Rightarrow A=\dfrac12+\dfrac1{2^2}\ +\,.\!.\!.+\ \dfrac1{2^{99}}\\\Rightarrow2A=2.\!\left(\dfrac12+\dfrac1{2^2}\ +\,.\!.\!.+\ \dfrac1{2^{99}}\right)\\\Rightarrow2A=1+\dfrac12\ +\,.\!.\!.+\ \dfrac1{2^{100}}\\\Rightarrow 2A-A=\left(1+\dfrac12\ +\,.\!.\!.+\ \dfrac1{2^{100}}\right)-\left(\dfrac12+\dfrac1{2^2}\ +\,.\!.\!.+\ \dfrac1{2^{99}}\right)\\\Rightarrow A=1-\dfrac1{2^{100}}<1\\\Rightarrow A<1$
    Vậy $A<1$.

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222-9+11+12:2*14+14 = ? ( )