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)`