Tính `A=1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+…+1/(1+2+3+…+2021)`.

Question

Tính `A=1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+…+1/(1+2+3+…+2021)`.

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Gabriella 3 tháng 2021-07-18T19:51:01+00:00 2 Answers 49 views 0

Answers ( )

    0
    2021-07-18T19:52:06+00:00

    Đáp án:2021/1011

     

    Giải thích các bước giải:Ta có 

    A=1+1/3+1/6+1/10+…..+1/2021×2022÷2

    A/2=1/2+1/6+1/12+1/20+…..+1/2021×2022

    A/2=1/1×2+1/2×3+1/3×4+…….+1/2021×2022

    A/2=1-1/2+1/2-1/3+1/3-1/4+….+1/2021-1/2022=1-1/2022=2021/2022

    A=2021/1011

     

    0
    2021-07-18T19:52:27+00:00

    Đáp án: `A=2021/1011`

     

    Giải thích các bước giải:

    `(1+2+3+…+2021=(2021+1)/(2).2021=1011.2021)`

     `A=1+(1)/(1+2)+(1)/(1+2+3)+(1)/(1+2+3+4)+…..+(1)/(1+2+3+…+2021)`

    `=1+(1)/(3)+(1)/(6)+(1)/(10)+….+(1)/(1011.2021)`

    `=1+(2)/(6)+(2)/(12)+(2)/(20)+…..+(2)/(2021.2022)`

    `=1+2.((1)/(2.3)+(1)/(3.4)+(1)/(4.5)+….+(1)/(2021.2022))`

    `=1+2.(1/2-1/3+1/3-1/4+1/4-1/5+……+1/2021-1/2022)`

    `=1+2.(1/2-1/2022)`

    `=1+1010/1011`

    `=1011/1011+1010/1011`

    `=2021/1011`

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