Ta có:

\(\frac{n\left(n+2\right)}{\left(n+1\right)^2}=1-\frac{1}{\left(n+1\right)^2}>1-\frac{1}{n\left(n+2\right)}=1+\frac{1}{2}.\left(\frac{1}{n+2}-\frac{1}{n}\right)\)

Thế vô bài toán ta được

\(B=\frac{2.4}{3^2}+\frac{4.6}{5^2}+...+\frac{200.202}{201^2}\)

\(>1+1+...+1+\frac{1}{2}.\left(\frac{1}{4}-\frac{1}{2}+\frac{1}{6}-\frac{1}{4}+...+\frac{1}{202}-\frac{1}{200}\right)\)

\(=100+\frac{1}{2}.\left(\frac{1}{202}-\frac{1}{2}\right)=\frac{10075}{101}>99,75\)

Ta có đánh giá sau:\(\frac{n\left(n+2\right)}{\left(n+1\right)^2}=1-\frac{1}{\left(n+1\right)^2}\)

\(>1-\frac{1}{x\left(x+2\right)}=1-\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)\)

Suy ra \(B=\frac{2\cdot4}{3^2}+\frac{4\cdot6}{5^2}+\frac{6\cdot8}{7^2}+...+\frac{200\cdot202}{201^2}\)

\(>1-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}\right)+1-\frac{1}{2}\left(\frac{1}{4}-\frac{1}{6}\right)+...+1-\frac{1}{2}\left(\frac{1}{200}-\frac{1}{202}\right)\)

\(=100-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{200}-\frac{1}{202}\right)\)

\(=100-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{202}\right)\)\(=100-\frac{1}{2}\cdot\frac{50}{101}\)

\(>100-\frac{1}{2}\cdot\frac{50}{100}=100-0,25=99,75\)

Tức là \(B>99,75\) 

v~ thành nhai lại rồi :V

khó quá

B = \(\frac{8}{9}+\frac{24}{25}+\frac{48}{49}+...+\frac{200.202}{201^2}=\left(1-\frac{1}{3^2}\right)+\left(1-\frac{1}{5^2}\right)+\left(1-\frac{1}{7^2}\right)+...+\left(1-\frac{1}{201^2}\right)\)

\(=\left(1+1+1+...+1\right)-\left(\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+...+\frac{1}{201^2}\right)\)

 \(=100-\left(\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+...+\frac{1}{201^2}\right)\)

Ta có Đặt \(C=\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+....+\frac{1}{201^2}\)\

\(< \frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{199.201}=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{199.201}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{199}-\frac{1}{201}\right)=\frac{1}{2}\left(1-\frac{1}{201}\right)=\frac{1}{2}.\frac{200}{201}=\frac{100}{201}< \frac{1}{2}\)

=> C < 1/2

=> B > 100 - 1/2

=> B > 99,5

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