\(B=\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}<\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}=\frac{6}{25}<\frac{6}{24}=\frac{1}{4}\)=>B<\(\frac{1}{4}\)(1)

\(B=\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{96}{576}=\frac{1}{6}\)=>B>\(\frac{1}{6}\)(2)

Từ (1)(2)=> \(\frac{1}{6} (đpcm)

\(B=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)

\(B=\frac{1}{2^2.1}+\frac{1}{2^2.2^2}+\frac{1}{3^2.2^2}+...+\frac{1}{50^2.2^2}\)

\(B=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)

\(B=\frac{1}{2^2}\left(1+\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{50.50}\right)\)

Ta có :

\(\frac{1}{2.2}< \frac{1}{1.2};\frac{1}{3.3}< \frac{1}{2.3};\frac{1}{4.4}< \frac{1}{3.4};...;\frac{1}{50.50}< \frac{1}{49.50}\)

\(\Rightarrow B< \frac{1}{2^2}\left(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)\)

Nhận xét : \(\frac{1}{1.2}< 1-\frac{1}{2};\frac{1}{2.3}< \frac{1}{2}-\frac{1}{3};...;\frac{1}{49.50}< \frac{1}{49}-\frac{1}{50}\)

\(\Rightarrow B=\frac{1}{2^2}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\)

\(B< \frac{1}{2^2}\left(1-\frac{1}{50}\right)\)

\(B< \frac{1}{4}.\frac{49}{50}< 1\)

\(B< \frac{49}{200}< \frac{1}{2}\)

\(\Rightarrow B< \frac{1}{2}\)