Ta có : \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)

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

\(< \frac{1}{4}.\left(1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{\left(n-1\right)n}\right)\)

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

\(=\frac{1}{4}.\left(2-\frac{1}{n}\right)\)

\(=\frac{1}{2}-\frac{1}{4n}< 1\)

Vậy A < 1

\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}.\)

\(A=\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+...+\frac{1}{4n^2}.\)

\(A=\frac{1}{4}\left(1+\frac{1}{4}+\frac{1}{9}+...+\frac{1}{n^2}\right)\)

\(A=\frac{1}{4}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\)

So sánh \(\frac{1}{2^2}< \frac{1}{1\cdot2};\frac{1}{3^2}< \frac{1}{2\cdot3};....\)

\(\Rightarrow A< \frac{1}{4}\left(1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{n\left(n-1\right)}\right)\)

\(\Rightarrow A< \frac{1}{4}\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{n-1}+\frac{1}{n-1}-\frac{1}{n}\right)\)

\(\Rightarrow A< \frac{1}{4}\left(1+1-\frac{1}{n}\right)\)

\(\Rightarrow A< \frac{1}{4}\left(2-\frac{1}{n}\right)\)

\(\Rightarrow A< \frac{1}{2}-\frac{1}{4n}\)

có \(\frac{1}{2}>\frac{1}{2}-\frac{1}{4n}\)

\(\Rightarrow A< \frac{1}{2}-\frac{1}{4n}< \frac{1}{2}\) mà \(\frac{1}{2}< 1\)

\(\Rightarrow A< 1\)

\(A<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=1-\frac{1}{n}<1\)

\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)

\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)

\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)

a) Ta có :

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\)

\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}=1-\frac{1}{n}< 1\)

\(\Rightarrow\)A < 1 

b) \(B=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)

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

vì \(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}< 2-\frac{1}{n}< 2\)

\(\Rightarrow B< \frac{1}{2^2}.2=\frac{1}{2}\)

cảm ơn nha!