1/3 – 2/3^2 + 3/3^3 – … – 3/3^100 So sánh với 3/16

Question

1/3 – 2/3^2 + 3/3^3 – … – 3/3^100
So sánh với 3/16

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Elliana 1 ngày 2021-09-10T12:32:01+00:00 2 Answers 1 views 0

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    0
    2021-09-10T12:33:10+00:00

    Giải thích các bước giải:

    Đặt `A = 1/3 – 2/3^2 + 3/3^3 – … – 3/3^100`

    `=> 3A = 1 – 2/3 + 3/3^2 + ….. – 3/3^99`

    `=> 3A + A = (1 – 2/3 + 3/3^2 + ….. – 3/3^99) + (1/3 – 2/3^2 + 3/3^3 – … – 3/3^100)`

    `=> 4A = 1 – 1/3 + 1/3^2 – 1/3^3 + …. – 1/3^99 + 3/3^100`

    `=> 4A = B – 3/3^100`

    Với `B = 1 – 1/3 + 1/3^2 – 1/3^3 + …. – 1/3^99`

    `=> 3B = 3 – 1 + 1/3 – 1/3^2 + … – 1/3^98`

    `=> 3B + B = (3 – 1 + 1/3 – 1/3^2 + … – 1/3^98)+( 1 – 1/3 + 1/3^2 – 1/3^3 + …. – 1/3^99)`

    `=> 4B = 3 – 1/3^99`

    `=> B = 3/4 – 1/(3^99 . 4) < 3/4`

    Vậy `4A < 3/4 – 3/3^100 < 3/4`

    `=> A<3/16 ( Đpcm)`

     

    0
    2021-09-10T12:33:25+00:00

    Giải thích các bước giải:

    Đặt:

    $A=\dfrac13-\dfrac2{3^2}+\dfrac3{3^3}-…-\dfrac3{3^{100}}$

    $\to 3A=1-\dfrac2{3}+\dfrac3{3^2}-…-\dfrac3{3^{99}}$

    $\to 3A+A=1-\dfrac13+\dfrac1{3^2}-\dfrac1{3^3}+…-\dfrac{1}{3^{99}}-\dfrac3{3^{100}}$

    $\to 4A=1-\dfrac13+\dfrac1{3^2}-\dfrac1{3^3}+…-\dfrac{1}{3^{99}}-\dfrac3{3^{100}}$

     Ta có:

    $B=1-\dfrac13+\dfrac1{3^2}-\dfrac1{3^3}+…-\dfrac{1}{3^{99}}$

    $\to 3B=3-1+\dfrac1{3}-\dfrac1{3^2}+…-\dfrac{1}{3^{98}}$

    $\to 3B+B=3-\dfrac1{3^{99}}$

    $\to 4B=3-\dfrac1{3^{99}}$

    $\to B=\dfrac34- \dfrac1{4\cdot 3^{99}}$

    $\to 4A=\dfrac34-\dfrac1{4\cdot 3^{99}}-\dfrac3{3^{100}}$

    $\to 4A<\dfrac34$

    $\to A<\dfrac3{16}$

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