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\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\)

\(=\frac{1}{2^2}\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)< \frac{1}{4}\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\right)\)

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

Ta có:\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)

\(=\frac{1}{4.4}+\frac{1}{4.9}+\frac{1}{4.16}+...+\frac{1}{4.n^2}\)

\(=\frac{1}{4}\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{n^2}\right)\)

\(Xét:\)

\(\frac{1}{2.2}< \frac{1}{1.2};\frac{1}{3.3}< \frac{1}{2.3};\frac{1}{4.4}< \frac{1}{3.4};\frac{1}{n.n}< \frac{1}{\left(n-1\right).n}...\)

\(Suyra:\)

\(P=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{n.n}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)

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

\(\Leftrightarrow P< 1-\frac{1}{n}< 1\)

\(\Leftrightarrow\frac{1}{4}.P< 1.\frac{1}{4}\)

\(\Leftrightarrow\frac{1}{4}\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{n^2}\right)< \frac{1}{4}\)

\(\Leftrightarrow\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\left(đpcm\right)\)

\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\).... \(+\frac{1}{\left(2n\right)^2}\)= \(\frac{1}{2^2}\). ( \(\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{n^2}\)) < \(\frac{1}{2^2}\)( \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).\left(n\right)}\)) = \(\frac{1}{2^2}\)( \(1-\frac{1}{n}\)) < \(\frac{1}{2^2}\).1 = \(\frac{1}{4}\)

\(\Rightarrow\)\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)< \(\frac{1}{4}\)

mình ko hiểu lắm

bn đâu có phải hotgirl đâu

Ta có:

\(\frac{1}{4^2}=\frac{1}{4.4}< \frac{1}{3.4}\)

\(\frac{1}{6^2}=\frac{1}{6.6}< \frac{1}{5.6}\)

\(\frac{1}{8^2}=\frac{1}{8.8}< \frac{1}{7.8}\)

\(...\)

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

Vậy \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)\(< \)\(\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+...+\frac{1}{1n.2n}\)

\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)\(< \)\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{1n}-\frac{1}{2n}\)

\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)\(< \)\(\frac{1}{3}+\left(\frac{-1}{4}+\frac{1}{4}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+...-\frac{1}{2n}\)

\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)\(< \)\(\frac{1}{3}-\frac{1}{2n}\)

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

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

Có:

\(\frac{1}{2^2}< \frac{1}{1.2}\)

\(\frac{1}{3^2}< \frac{1}{2.3}\)

\(...\)

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

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

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1-\frac{1}{n}< 1\)

\(\Rightarrow A< \frac{1}{2^2}.1=\frac{1}{4}\)

Đặt A= \(\frac{1}{4^2}\) + \(\frac{1}{6^2}\) + \(\frac{1}{8^2}\) +...+ \(\frac{1}{\left(2n\right)^2}\)

A= \(\frac{1}{2^2.2^2}\) + \(\frac{1}{2^2.3^2}\) +...+ \(\frac{1}{2^2.n^2}\)

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

A< \(\frac{1}{2^2}\) . ( \(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) + \(\frac{1}{3.4}\) +...+ \(\frac{1}{\left(n-1\right)n}\)

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

A< \(\frac{1}{4}\) . (1-\(\frac{1}{n}\)) = \(\frac{1}{4}\) - \(\frac{1}{4n}\) <\(\frac{1}{4}\) => A <\(\frac{1}{4}\)