a ) \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}=\frac{1}{4}\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\)

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

b )

\(B=\frac{1}{3^2}+\frac{1}{5^2}+...+\frac{1}{\left(2n+1\right)^2}< \frac{1}{3^2-1}+\frac{1}{5^2-1}+...+\frac{1}{\left(2n+1\right)^2-1}\)

\(=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2n\left(2n+2\right)}\)

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

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

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 !!! ^_^

chứng minh bài toán theo cách quy nạp toán học.  

Với n=2 suy ra:\(\frac{1}{3}+\frac{1}{4}>\frac{13}{14}\left(TM\right)\)

Giả sử bài toán trên đúng với mọi n=k,ta cần chứng minh nó đúng với n=k+1,tức là:

\(S_k=\frac{1}{k+2}+\frac{1}{k+3}+\frac{1}{k+4}+....+\frac{1}{2\left(k+1\right)}>\frac{13}{14}\)

Thật vậy:

\(\frac{1}{k+2}+\frac{1}{k+3}+...+\frac{1}{2\left(k+1\right)}\)

\(=\frac{1}{k+1}+\frac{1}{k+2}+....+\frac{1}{2k}+\frac{1}{2k+1}+\frac{1}{2k+2}-\frac{1}{k+1}\)

\(=S_k+\frac{1}{2k+1}+\frac{1}{2k+2}-\frac{1}{k+1}\)

\(>\frac{13}{14}+\frac{2k+2}{2\left(k+1\right)\left(2k+1\right)}+\frac{2k+1}{2\left(k+1\right)\left(2k+1\right)}-\frac{2\left(2k+1\right)}{2\left(k+1\right)\left(2k+1\right)}\)

\(=\frac{13}{14}+\frac{2\left(k+1\right)+2k+1-2\left(2k+1\right)}{2\left(k+1\right)\left(2k+1\right)}\)

để dễ hiểu,,mik xin viết thêm nha(không phải để kiếm điểm,có người nhờ nên mới thế này:))

\(\frac{13}{14}+\frac{2\left(k+1\right)+2k+1-2\left(2k+1\right)}{2\left(k+1\right)\left(2k+1\right)}\)

\(=\frac{13}{14}+\frac{1}{2\left(k+1\right)\left(2k+1\right)}>\frac{13}{14}\left(k>1\right)\)

\(\Rightarrow S_{k+1}>\frac{13}{14}\)

\(\Rightarrow S_k>\frac{13}{14}\)

Phép chứng minh hoàn tất_._

 \(S=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}>\frac{13}{24}\left(1\right)\)

Với : \(n=2\), suy ra : \(\frac{1}{3}+\frac{1}{4}=\frac{7}{12}>\frac{13}{24}\left(TM\right)\)

Giả sử : (1) đúng với : \(n=k\left(k>1\right)\), tức là :

\(S_k=\frac{1}{k+1}+\frac{1}{k+2}+...+\frac{1}{2k}>\frac{13}{24}\)( giả thiết quy nạp )

Ta cần c/m : (1) đúng với : \(n=k+1\), tức là cần chứng minh : 

\(S_{k+1}=\frac{1}{k+2}+\frac{1}{k+3}+...+\frac{1}{2k}+\frac{1}{2k+1}+\frac{1}{2\left(k+1\right)}>\frac{13}{24}\)

Thật vậy : \(S_{k+1}=\frac{1}{k+2}+\frac{1}{k+3}+...+\frac{1}{2k}+\frac{1}{2k+1}+\frac{1}{2\left(k+1\right)}\)

\(=\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}+...+\frac{1}{2k}+\frac{1}{2k+1}+\frac{1}{2k+2}-\frac{1}{k+1}\)

\(=S_k+\frac{1}{2k+1}+\frac{1}{2k+2}-\frac{1}{k+1}>\frac{13}{24}+\frac{1}{2k+1}+\frac{1}{2k+2}-\frac{1}{k+1}\)

\(=\frac{13}{24}+\frac{2k+2}{2\left(k+1\right)\left(2k+1\right)}+\frac{2k+1}{2\left(k+1\right)\left(2k+1\right)}-\frac{2\left(2k+1\right)}{2\left(k+1\right)\left(2k+1\right)}\)

\(=\frac{13}{24}+\frac{2\left(k+1\right)+2k+1-2\left(2k+1\right)}{2\left(k+1\right)\left(2k+1\right)}\)

\(=\frac{13}{24}+\frac{2k+2+2k+1-4k-2}{2\left(k+1\right)\left(2k+1\right)}\)

\(=\frac{13}{24}+\frac{1}{2\left(k+1\right)\left(2k+1\right)}>\frac{13}{24}\left(k>1\right)\)

Vậy : \(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}>\frac{13}{24}\)đúng với mọi \(n>1\)

Đặt \(A=\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+...+\frac{1}{\left(2n+1\right)^2}\)

Ta có : \(\left(2n+1\right)^2=4n^2+4n+1>4n^2+4n\Leftrightarrow\left(2n+1\right)^2>2n\left(2n+2\right)\)\(\Leftrightarrow\frac{1}{\left(2n+1\right)^2}< \frac{1}{2n\left(2n+2\right)}\)

Mà \(\hept{\begin{cases}\frac{1}{3^2}< \frac{1}{2.4}\\\frac{1}{5^2}< \frac{1}{4.6}\\\frac{1}{7^2}< \frac{1}{6.8}\end{cases}}\)

\(...............\)

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

\(\Rightarrow\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+...+\frac{1}{\left(2n+1\right)^2}< \frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2n\left(2n+2\right)}=B\)

\(=\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{2n+2-2n}{2n\left(2n+2\right)}\)

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2n}-\frac{1}{2n+2}\)

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

\(\Rightarrow A< B< \frac{1}{4}\Rightarrow A< \frac{1}{4}\) hay đpcm