Question

chứng tỏ rằng : \(\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{2n^2}< \frac{1}{4}\)(nϵN,n≥2)

Answer 1

Đặt \(A=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{2n^2}\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{\left(2n-2\right).2n}\right)\)

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

\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2n}\right)\)

\(\Rightarrow A=\frac{1}{2}.\frac{1}{2}-\frac{1}{2}.\frac{1}{2n}\)

\(\Rightarrow A=\frac{1}{4}-\frac{1}{4n}\)

Vì \(\frac{1}{4}-\frac{1}{4n}< \frac{1}{4}.\)

\(\Rightarrow A< \frac{1}{4}\left(đpcm\right)\left(n\in N;n\ge2\right).\)

Chúc bạn học tốt!