A=1/3+2/3^2+3/3^3+4/3^4+….+100/3^100.Chứng tỏ rằng A<3/4

A=1/3+2/3^2+3/3^3+4/3^4+….+100/3^100.Chứng tỏ rằng A<3/4

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  1. Đáp án:

    `A<3/4`

    Giải thích các bước giải:

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

    `=>3A=3(1/3+2/3^2+3/3^3+4/3^4+…+99/3^99+100/3^100)`

    `=>3A=1/(3÷3)+2/(3^2÷3)+3/(3^3÷3)+4/(3^4÷3)+…+99/(3^99÷3)+100/(3^100÷3)`

    `=>3A=1+2/3+3/3^2+4/3^3+…+99/3^98+100/3^99`

    `=>3A-A=(1+2/3+3/3^2+4/3^3+…+99/3^98+100/3^99)-(1/3+2/3^2+3/3^3+4/3^4+…+99/3^99+100/3^100)`

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

    `=>6A=3(1+1/3+1/3^2+1/3^3+…+1/3^99+100/3^100)`

    `=>6A=3+1/(3÷3)+1/(3^2÷3)+1/(3^3÷3)+…+1/(3^99÷3)+100/(3^100÷3)`

    `=>6A=3+1+1/3+1/3^2+…+1/3^98+100/(3^99)`

    `=>6A-2A=(3+1+1/3+1/3^2+…+1/3^98+100/(3^99))-(1+1/3+1/3^2+1/3^3+…+1/3^99+100/3^100)`

    `=>4A=3-100/3^99+100/3^100`

    `=>`$A=\dfrac{3-\dfrac{100}{3^{100}}+\dfrac{101}{3^{101}}}{4}$

    `=>`$A=\dfrac{3}{4}-\dfrac{\dfrac{100}{3^{100}}+\dfrac{101}{3^{101}}}{4}<\dfrac34$

    Vậy `A<3/4`.

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