bài này tui bít làm nhưng dài lắm

Cố gắng làm hộ mình với !

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

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

Ta có:\(\frac{1}{2^2}\)<\(\frac{1}{1\cdot2}\)

         \(\frac{1}{3^2}\)<\(\frac{1}{2\cdot3}\)

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

=>\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{n^2}\)<\(\frac{1}{1\cdot2}\)+\(\frac{1}{2\cdot3}\)+...+\(\frac{1}{\left(n-1\right)\cdot n}\)

=>A<1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+...+\(\frac{1}{n-1}\)--\(\frac{1}{n}\)
=>\(\frac{1}{2^2}\)*A<\(\frac{1}{2^2}\)(1--\(\frac{1}{n}\))

=>\(\frac{1}{2^2}\)*A<\(\frac{1}{4}\)(1--\(\frac{1}{n}\))

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

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

đặt A=1/2^2+1/4^2+1/6^2+.....+1/(2n)^2

ta có :

A=1/2^2 +1/2^2(1/2^2+1/3^2+1/4^2+.....+1/n^2)

A<1/2^2+1/2^2(1/1.2+1/2.3+...+1/(n-1)n)

=1/2^2+1/2^2(1-1/2+1/2-1/3+....+1/(n-1)-1/n)

=1/2^2+1/2^2(1-1/n)

<1/2^2+1/2^2.1=1/2<3/4

vậy A<3/4

mình đồng ý với bạn witch roses

ta có 

\(\frac{1}{2^2}

Ta có:

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

\(=\frac{1}{2^2.2^2}+\frac{1}{2^2.3^2}+\frac{1}{2^2.4^2}+...+\frac{1}{2^2.n^2}\)

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

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

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

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

\(=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{n.n}\)

Vì: \(\frac{1}{2.2}

kho qua minh moi lop 5

What? lớp 5 mà học cả số mũ?? thời nay bọn trẻ con học trâu thật!

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

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

Ta co =>A=1/2^2(1/2^2+1/2^3+1/2^4+...+1/n^2)

                                ...................

 =>1/(1.2)+1/(2.3)+...+1/(n-1).n=1/1-1/n=1-1/n

=>A=1/4(1-1/n)

Mà 1-1/n>1

=>A=1/4(1-1/n)<1/4.1+1/4

Vay A<1/4

\(\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{n^2}

trên là 2n ở dưới lại là n