CMR: A=1/3-2/3^2+3/3^3-4/3^4+…+99/3^99-100/3^100

By aikhanh

CMR: A=1/3-2/3^2+3/3^3-4/3^4+…+99/3^99-100/3^100

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  1. Đáp án:

     

    Giải thích các bước giải:

    Ta có:` A=1/3 – 2/3^2+3/3^3 – 4/3^4+ … – 100/3^100`
    `=>3A=1 -2/3 +3/3^2 – 4/3^3+ … – 100/3^99`
    `=>4A=A+3A=1-1/3+1/3^2-1/3^3+…-1/3^99 – 100/3^100`
    `=>12A=3.4A=3-1+1/3-1/3^2+…-1/3^98 – 100/3^99`
    `=>16A=12A+4A=3-1/3^99-100/3^99-100/3^1…`
    `<=> 16A=3-101/3^99-100/3^100`
    `<=> A=3/10-(101/3^99+100/3^100)/16 < 3/10`
    Suy ra `A<3/10`

    Trả lời

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