Question

[#𝒌𝒂𝒎𝒊ㅤ♪]

Chứng tỏ rằng :\(B=1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{3^2}-...-\frac{1}{2004^2}>\frac{1}{2004}\)

Answer 1

[I don't need you]

Đề: cmr: B = 1 - 1/2² - 1/3² - 1/4² -...-1/2004² > 1/2004 ( bn có ghi nhầm đề ko z)

Bài làm

ta có: \(\frac{1}{2^2}>\frac{1}{1.2};\frac{1}{3^2}>\frac{1}{2.3};\frac{1}{4^2}>\frac{1}{3.4};...;\frac{1}{2004^2}>\frac{1}{2003.2004}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2004^2}>\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2003.2004}\)= 2003/2004

\(\Rightarrow B=1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2004^2}\right)>1-\frac{2003}{2004}=\frac{1}{2004}\)

=> đpcm

Answer 2

[tth_new]

@I don't need you: Hey \(\frac{1}{2^2}>\frac{1}{1.2}\Leftrightarrow0.25>0.5?!?\)

\(B=1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{2004^2}>\frac{1}{2004}\)

          Giải

Có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{2004^2}< \frac{1}{2003.2004}\)

\(B=1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2004^2}\right)\)

\(>1-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2003.2004}\right)\)

\(=1-\left(1-\frac{1}{2004}\right)=\frac{1}{2004}\) (đpcm)