Cho biểu thức $C$ = $\frac{1}{3}$ – $\frac{2}{3^{2}}$ + $\frac{3}{3^{3}}$ – $\frac{4}{3^{4}}$ + …….. + $\frac{99}{3^{99}}$ – $\frac{100}{3^{100}}$

Cho biểu thức $C$ = $\frac{1}{3}$ – $\frac{2}{3^{2}}$ + $\frac{3}{3^{3}}$ – $\frac{4}{3^{4}}$ + …….. + $\frac{99}{3^{99}}$ – $\frac{100}{3^{100}}$
Chứng minh rằng $C$ < $\frac{3}{16}$

0 bình luận về “Cho biểu thức $C$ = $\frac{1}{3}$ – $\frac{2}{3^{2}}$ + $\frac{3}{3^{3}}$ – $\frac{4}{3^{4}}$ + …….. + $\frac{99}{3^{99}}$ – $\frac{100}{3^{100}}$”

  1. `C= 1/3 – 2/3^2 + 3/3^3 – 4/3^4+…+99/3^99 – 100/3^100`

    `3C= 3(1/3 – 2/3^2 + 3/3^3 – 4/3^4+…+99/3^99 – 100/3^100)`

    `3C= 1 – 2/3 + 3/3^2 – 4/3^3 +…+ 99/3^98 – 100/3^99`

    `3C+C = 1-2/3 + 3/3^2 – 4/3^3 +…+99/3^98 – 100/3^99 +1/3 – 2/3^2 + 3/3^3 – 5/3^4 +…+99/3^99 – 100/3^100`

    `4C= 1- 1/3 + 1/3^2 – 1/3^3+….+1/3^98 – 1/3^99`

    Đặt `A= 1 – 1/3 + 1/3^2 -1/3^3+…-1/3^99`

    `1/3 A = 1/3( 1 – 1/3 + 1/3^2 -1/3^3+…-1/3^99)`

    `1/3A = 1/3 – 1/3^2 + 1/3^3-….-1/3^100`

    `A+ 1/3 A = 1 – 1/3 + 1/3^2 -1/3^3+…+1/3^98 – 1/3^99 + 1/3 -1/3^2 + 1/3^3 -….-1/3^100`

    `4/3 A= 1- 1/3^100`

    `A = (1-1/3^100) :4/3`

    `A= 3/4 – 1/(3^99. 4)`

    `=> 4C = (3/4 – 1/(3^99. 4))`

    `=> C = (3/4 – 1/(3^99.4)) : 4`

    `C= (3/4 – 1/(3^99. 4)) . 1/4`

    `C= 3/16 – 1/(3^99. 16) < 3/16`

    Vậy `C < 3/16`

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