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)
Chứng Minh Rằng: 1/4^2+1/6^2+1/8^2+...+1/(2.n)^2<1/4
