Cho: A=1/3^2+3/3^4+5/3^6+…+99/3^100

Chứng minh: A<5/32

Đáp án`+`Giải thích các bước giải:

 Ta có :`A=1/3^2+3/3^4+5/3^6+...+99/3^100`

`=>9A=1+3/3^2+5/3^4+...+99/3^98`

`=>9A-A=1+2/3^2+2/3^4+...+2/3^98-99/3^100`

`=>8A=2(1/3^2+1/3^4+...+1/3^98)+1-99/3^100`

Đặt `B=(1/3^2+1/3^4+...+1/3^98)`

`=>9B=1+1/3^2+...+1/3^96`

`=>9B-B=1-1/3^98`

`=>B=(1-1/3^98)/8`

Khi đó : `8A=(1-1/3^98)/4+1-99/3^100`

`=>A=(1-1/3^98)/32+1/8-99/(8.3^100)`

`=>A=(1/32+1/8)-(1/(32.3^98)+99/(8.3^100))`

`=>A=5/32-X` với `X>0`

`=>A<5/32`

Vậy `A<5/32` `(đpcm)`