khó quá

Ta có : 

\(B=\frac{8}{9}+\frac{24}{25}+...+\frac{200.202}{201^2}\)

\(B=\frac{8}{3^2}+\frac{24}{5^2}+...+\frac{200.202}{201^2}\)

\(B=\frac{3^2-1}{3^2}+\frac{5^2-1}{5^2}+...+\frac{201^2-1}{201^2}\)

\(B=\frac{3^2}{3^2}-\frac{1}{3^2}+\frac{5^2}{5^2}-\frac{1}{5^2}+...+\frac{201^2}{201^2}-\frac{1}{201^2}\)

\(B=1-\frac{1}{3^2}+1-\frac{1}{5^2}+...+1-\frac{1}{201^2}\)

\(B=\left(1+1+...+1\right)+\left(-\frac{1}{3^2}-\frac{1}{5^2}-...-\frac{1}{201^2}\right)\)

\(B=100-\left(\frac{1}{3^2}+\frac{1}{5^2}+...+\frac{1}{201^2}\right)\)

Lại có : 

\(\frac{1}{3^2}+\frac{1}{5^2}+...+\frac{1}{201^2}>\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{201.203}\)

\(\Leftrightarrow\)\(\frac{2}{3^2}+\frac{2}{5^2}+...+\frac{2}{201^2}>\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{201.203}\)

\(\Leftrightarrow\)\(\frac{2}{3^2}+\frac{2}{5^2}+...+\frac{2}{201^2}>\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{201}-\frac{1}{203}\)

\(\Leftrightarrow\)\(\frac{2}{3^2}+\frac{2}{5^2}+...+\frac{2}{201^2}>\frac{1}{3}-\frac{1}{203}\)

\(\Leftrightarrow\)\(\frac{2}{3^2}+\frac{2}{5^2}+...+\frac{2}{201^2}>\frac{200}{609}\)

Suy ra : \(2B=200-\left(\frac{2}{3^2}+\frac{2}{5^2}+...+\frac{2}{201^2}\right)>200-\frac{200}{609}\)

\(\Leftrightarrow\)\(B>100-\frac{100}{609}\)

\(\Leftrightarrow\)\(B>\frac{60800}{609}=99,\left(835...99\right)>99,75\)

Vậy \(B>99,75\)

Chúc bạn học tốt ~ 

Bạn có thể giải thích tại sao lại \(2B=200-\left(\frac{2}{3^2}+\frac{2}{5^2}+...+\frac{2}{201^2}\right)>200-\frac{200}{609}\)  từ đoạn đó xuống dưới đc ko