$\dfrac{1}{1.2.3}$ + $\dfrac{1}{2.3.4}$ + $\dfrac{1}{3.4.5}$ +…+ $\dfrac{1}{100.101.102}$

$\dfrac{1}{1.2.3}$ + $\dfrac{1}{2.3.4}$ + $\dfrac{1}{3.4.5}$ +…+ $\dfrac{1}{100.101.102}$

0 bình luận về “$\dfrac{1}{1.2.3}$ + $\dfrac{1}{2.3.4}$ + $\dfrac{1}{3.4.5}$ +…+ $\dfrac{1}{100.101.102}$”

  1. Đáp án+Giải thích các bước giải:

    $Đặt A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+…+\dfrac{1}{100.101.102}\\=>2A=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+…+\dfrac{2}{100.101.102}\\=\dfrac1{1.2}-\dfrac1{2.3}+\dfrac1{2.3}-…+\dfrac1{100.101}-\dfrac1{101.102}\\=\dfrac1{2}-\dfrac1{101.102}\\=\dfrac{2575}{5151}\\=>A=\dfrac{\dfrac{2575}{5151}}{2}=\dfrac{2575}{10302}$

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  2. Đáp án:

    ` 1/(1.2.3)+1/(2.3.4)+1/(3.4.5)+…+1/(100.101.102)`

    `= 1/2(2/(1.2.3)+2/(2.3.4)+2/(3.4.5)+…+2/(100.101.102))`

    `= 1/2((3-1)/(1.2.3)+(4-2)/(2.3.4)+(5-3)/(3.4.5)+…+(102-100)/(100.101.102)`

    `=1/2(1/(1.2)-1/(2.3)+1/(2.3)-1/(3.4)+1/(3.4)-1/(4.5)+…+1/(100.101)-1/(101.102))`

    `=1/2(1/2-1/10302)`

    `=1/2 . 2575/5151`

    `=2575/10302`

     

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