HEEEEELP Cho biểu thức: C = $\frac{1}{3}$ – $\frac{2}{3^{2}}$ + $\frac{3}{3^3}$ – $\frac{3}{3^4}$ + … + $\frac{99}{3^{99}}$ – $\frac{100}{3^{10

Bởi

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

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  1. Giải thích các bước giải:

     Ta có: C=`1/3`-`2/2^3`+…+`99/3^99`-`100/3^100`

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

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

    `4`C=`1`-`2/3+`3/3^2`-`4/3^3`+…+`9/3^98-`100/3^99`+`1/3`-`2/3^2`+…+`9/9^99`-`100/3^100`

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

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

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

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

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

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

    `4`A= 3+`(1/3-1/3)`+`(-1/3^2+1/3^2)`+…+`(-1/3^90+1/3^99)`-`1/3^99`

    `4`A= `3`- `1/3^99`

    => A=`3/4`_`1/3^99`

    Vậy `4`C=`(3/4-1/4.3^99-100/3^100)`

    C= `1/4`.`(3/4-1/4^4.3^99-100/3^100)`

    C= `3/16`-`(1/4^2.3^99+25/3^100)`

    Vì `(1/4^2.3^99+25/3^100)` >`0`  => C = `3/16`-`(1/4^2.3^99+25/3^100)` < `3/16` `(đpcm)`

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