xét tổng T=2/2+3/2^2+4/2^3+…+2019/2^2018+2020/2^2019 so sánh T với 3

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xét tổng T=2/2+3/2^2+4/2^3+…+2019/2^2018+2020/2^2019 so sánh T với 3

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Iris 3 tháng 2021-09-13T20:48:42+00:00 1 Answers 21 views 0

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    2021-09-13T20:50:35+00:00

    Đáp án: $T<3$

    Giải thích các bước giải:

    Ta có:

    $T=\dfrac22+\dfrac3{2^2}+\dfrac4{2^3}+…+\dfrac{2019}{2^{2018}}+\dfrac{2020}{2^{2019}}$

    $\to 2T=2+\dfrac3{2}+\dfrac4{2^2}+…+\dfrac{2019}{2^{2017}}+\dfrac{2020}{2^{2018}}$

    $\to 2T-T=2+\dfrac12+\dfrac1{2^2}+…+\dfrac{1}{2^{2018}}-\dfrac{2020}{2^{2019}}$

    $\to T=2+\dfrac12+\dfrac1{2^2}+…+\dfrac{1}{2^{2018}}-\dfrac{2020}{2^{2019}}$

    Mà $A=\dfrac12+\dfrac1{2^2}+…+\dfrac{1}{2^{2018}}$

    $\to 2A=1+\dfrac12+…+\dfrac1{2^{2017}}$

    $\to 2A-A=1-\dfrac{1}{2^{2018}}$

    $\to A=1-\dfrac{1}{2^{2018}}$

    $\to T=2+1-\dfrac{1}{2^{2018}}-\dfrac{2020}{2^{2019}}$

    $\to T=3-\dfrac{1}{2^{2018}}-\dfrac{2020}{2^{2019}}$

    $\to T<3$

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