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

Toán Lớp 7: Chứng minh rằng: 1/2^2 + 1/3^2 + 1/4^2 + … + 1/2021^2 < 3/4

Toán Lớp 7: Chứng minh rằng: 1/2^2 + 1/3^2 + 1/4^2 + … + 1/2021^2 < 3/4

Comments ( 2 )

  1. Giải đáp:
     
    Lời giải và giải thích chi tiết:
    A=1/2^2+1/3^2+1/4^2+…+1/2021^2
    =>A=1/4+1/3.3+1/4.4+…+1/2021.2021
    =>A<1/4+1/2.3+1/3.4+…+1/2020.2021
    =>A<1/4+1/2-1/3+1/3-1/4+…+1/2020-1/2021
    =>A<1/4+1/2-1/2021
    =>A<3/4-1/2021<3/4
    =>A<3/4
    Vậy biểu thức có giá trị bé hơn 3/4.

  2. Giải đáp:
    $\\$
    Đặt A=1/2^2 + 1/3^2 + 1/4^2 + … + 1/2021^2
    -> A = 1/4 + 1/3^2 + 1/4^2 + … + 1/2021^2
    $\bullet$ 1/3^2 + 1/4^2 + … + 1/2021^2
    Vì 3.3 > 2.3
    -> 1/3^2 < 1/(2.3)
    Tương tự có :
    1/4^2 < 1/(3.4)
    ………..
    1/2021^2 < 1/(2020.2021)
    Cộng theo vế ta được :
    -> 1/3^2 +1/4^2 + … + 1/2021^2 < 1/(2.3) + 1/(3.4) + … + 1/(2020 . 2021)
    -> 1/3^2 + 1/4^2 + … + 1/2021^2 < 1/2 – 1/3 +1/3 – 1/4 + … + 1/2020 – 1/2021
    -> 1/3^2 + 1/4^2 + … + 1/2021^2 < 1/2 + (-1/3 + 1/4) + … + (-1/2020 + 1/2020) – 1/2021
    -> 1/3^2 + 1/4^2 + … + 1/2021^2 < 1/2 – 1/2021
    -> 1/4 + 1/3^2 + 1/4^2 + … + 1/2021^2 < 1/4 + 1/2 – 1/2021
    -> A < 3/4 – 1/2021
    Ta thấy : 3/4 – 1/2021 < 3/4
    -> A < 3/4 – 1/2021 < 3/4
    -> A<3/4 (đpcm)
     

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

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