Đáp án:
`B=1-1/2^2-1/3^2-....-1/2004^2`
`=1-(1/2^2+1/3^2+....+1/2004^2)`
Ta có : `1/2^2>1/(1.2) ; 1/3^2>1/(2.3) ; .... ; 1/2004^2>1/(2003.2004)`
`=> B>1-(1/(1.2)+1/(2.3)+...+1/(2003.2004))`
`B>1-(1/1-1/2+1/2-1/3+...+1/2003-1/2004)`
`B>1-(1/1-1/2004)=1-1+1/2004=1/2004`
Vậy `B>1/2004`