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\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{\left(2n-2\right)2n}\)
\(\Rightarrow\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}< \frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{\left(2n-2\right)2n}\)\(.\frac{1}{2}\) Ta gọi là A
\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{\left(2n-2\right)2n}\right)\)
\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2n}\right)=\frac{1}{2}.\frac{1}{2}-\frac{1}{2}.\frac{1}{2n}=\frac{1}{4}-\frac{1}{2n.2}\)
\(\Rightarrow M< \frac{1}{4}-\frac{1}{2n.2}< \frac{1}{4}\)
\(\Rightarrow M< \frac{1}{4}\left(Đpcm\right)\)
\(\)
\(N=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)
\(N=\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)< \frac{1}{2^2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{...1}{\left(n-1\right).n}\right)\)
\(N< \frac{1}{4}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)
\(N< \frac{1}{4}.\left(1-\frac{1}{n}\right)< \frac{1}{4}.1=\frac{1}{4}\)
=> \(N< \frac{1}{4}\)(đpcm)
\(\begin{equation} x = a_0 + \cfrac{1}{740_1 + \cfrac{1}{897654_2 + \cfrac{1}{672_3 + \cfrac{1}{100_4} } } } \end{equation}\)
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
........
\(\frac{1}{n^2}< \frac{1}{\left(n-1\right)n}\)
=> \(A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n\left(n-1\right)}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}=1-\frac{1}{n}< 1\)
Đpcm
Số shạng tổng quát là \(\frac{1}{\left(2n\right)^2}.\) mới phải đó bạn ơi.
\(A=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{2}{\left(2n\right)^2}< \frac{1}{2}\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{\left(2n-1\right)2n}\right)=.\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2n-1}-\frac{1}{2n}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2n}\right)=\frac{1}{4}-\frac{1}{4n}< \frac{1}{4}.\)
Vậy \(A< \frac{1}{4}\)
Đặt \(A=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)
\(\Rightarrow A=\frac{1}{2^2}\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)\)
\(\Rightarrow A< \frac{1}{2^2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\right)\)
\(\Rightarrow A< \frac{1}{4}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\right)\)
\(\Rightarrow A< \frac{1}{4}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{\left(n-1\right)}-\frac{1}{n}\right)\)
\(\Rightarrow A< \frac{1}{4}\left(1-\frac{1}{n}\right)\)
\(\Rightarrow A< \frac{1}{4}-\frac{1}{4n}< \frac{1}{4}\)
Vậy \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\left(đpcm\right)\)