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Ta có :
\(N=\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)
\(N=\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\)
Ta thấy : \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
.......
\(\frac{1}{n^2}< \frac{1}{\left(n-1\right).n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right).n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1-\frac{1}{n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1\)
\(\Rightarrow\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)< 1.\frac{1}{2^2}\)
\(\Rightarrow N< \frac{1}{4}\)(ĐPCM)
Ủng hộ mk nha !!! ^_^
\(N=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)
\(N=\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)< \frac{1}{2^2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{...1}{\left(n-1\right).n}\right)\)
\(N< \frac{1}{4}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)
\(N< \frac{1}{4}.\left(1-\frac{1}{n}\right)< \frac{1}{4}.1=\frac{1}{4}\)
=> \(N< \frac{1}{4}\)(đpcm)
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 \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\left(đpcm\right)\)
Đặt \(A=\frac{3}{9.14}+\frac{3}{14.19}+.......+\frac{3}{\left(5n-1\right)\left(5n+4\right)}\)
\(5A=\frac{15}{9.14}+\frac{15}{14.19}+.....+\frac{15}{\left(5n-1\right)\left(5n+4\right)}\)
\(5A=3.\left(\frac{5}{9.14}+\frac{5}{14.19}+......+\frac{5}{\left(5n-1\right)\left(5n+4\right)}\right)\)
\(5A=3.\left(\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+.....+\frac{1}{5n-1}-\frac{1}{5n+4}\right)\)
\(5A=3.\left(\frac{1}{9}-\frac{1}{5n+4}\right)\)
\(5A=\frac{1}{3}-\frac{1}{5n+4}\)
=> \(5A<\frac{1}{3}\)
=> \(A<\frac{1}{3}:5\)
hay \(A<\frac{1}{15}\) \(\left(đpcm\right)\)
Nhớ nhé bạn
\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}\) ta có :
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(A< 1-\frac{1}{2010}=\frac{2009}{2010}< 1\)
\(\Rightarrow\)\(A< 1\) ( đpcm )
Vậy \(A< 1\)
Chúc bạn học tốt ~
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
........
\(\frac{1}{n^2}< \frac{1}{\left(n-1\right)n}\)
=> \(A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n\left(n-1\right)}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}=1-\frac{1}{n}< 1\)
Đpcm