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

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

=>A<1/4(1-1/2+1/2-1/3+...+1/n-1-1/n)

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

bn rất tốt nhưng mk rất tiếc phải ns câu này

: mấy bn ấy qa OLM cổ chơi hết rùi

Ta có: \(B=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{100^2}< \frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+...+\frac{1}{98\cdot100}\)

                                                         \(B< \frac{1}{2}\cdot\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{98\cdot100}\right)\)

                                                         \(B< \frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)

                                                         \(B< \frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{100}\right)\)

                                                         \(B< \frac{1}{4}-\frac{1}{200}< \frac{1}{4}\)

     \(\Rightarrow B< \frac{1}{4}\)

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

Ta có:
N=1/(2.2)^2+1/(2.3)^2+1/(2.4)^2+....+1/(2.n)^2
N=1/2^2.2^2+1/2^2.3^2+1/2^2.4^2+....+1/2^2.n^2
N=1/2^2.(1/2^2+1/3^2+1/4^2+...+1/n^2)
<1/4.(1/1.2+1/2.3+1/3.4+...1/(n-1).n)
=1/4.(1-1/n)<1/4.1=1/4 (vì n thuộc N,n lớn hơn hoặc bằng 2)

Ta có 3.5<4.4=4^2

5.7<6^2

...

(2n-1)(2n+1)<4n^2

Do vậy 1/4^2+1/6^2+....+1/4n^2<1/3.5+1/5.7+....+1/(2n-1)(2n+1)

=1/2[1/3-1/5+1/5-.....+1/(2n-1)-1/(2n+1)]

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

=1/6-1/2(2n+1)<1/4. Vậy ta có đpcm

Ta có 4^2>3.5

6^2>5.7

...

(2n-1)(2n+1)<4n^2

Do vậy 1/4^2+1/6^2+....+1/4n^2<1/3.5+1/5.7+...+1/(2n-1)(2n+1)

=1/2(1/3-1/5+1/5-...+1/2n-1-1/2n+1)

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

=1/6-1/2(2n+1)<1/4 (đpcm

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

=>\(S< =\dfrac{1}{4}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\right)\)

=>\(S< =\dfrac{1}{4}\cdot\left(1-\dfrac{1}{n}\right)=\dfrac{1}{4}\cdot\dfrac{n-1}{n}< =\dfrac{1}{4}\)