A=1/4+1/9+1/16+...+1/4012009 . CMR:1/3<A<1

Đáp án + Giải thích các bước giải:

`A=1/4+1/9+1/16+...+1/4012009`

`=>A=1/2^2+1/3^2+1/4^2+...+1/2003^2`

Ta có: `A=1/2^2+1/3^2+1/4^2+...+1/2003^2`

`=>A>1/2.3+1/3.4+1/4.5+...+1/2003.2004`

`=>A>1/2-1/3+1/3-1/4+1/4-1/5+...+1/2003-1/2004`

`=>A>1/2-1/2004=1002/2004-1/2004=1001/2004>668/2004=1/3(1)`

Lại có: `A=1/2^2+1/3^2+1/4^2+...+1/2003^2`

`=>A<1/1.2+1/2.3+1/3.4+...+1/2002.2003`

`=>A<1-1/2+1/2-1/3+1/3-1/4+...+1/2002-1/2003`

`=>A<1-1/2003<1(2)`

Từ `(1),(2)=>1/3<A<1`

Vậy `1/3<A<1`