Question

Chứng minh rằng\(\dfrac{1}{4^2} + \dfrac{1}{6^2} + \dfrac{1}{8^2} + ....... + \dfrac{1}{2n^2}< \dfrac{1}{4}\)

Answer 1

Ta có:

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

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

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{12}+...+\frac{1}{2n-2}-\frac{1}{2n}\right)\)

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

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

Vì \(\frac{1}{4}-\frac{1}{2n.2}< \frac{1}{4}\)

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

Vậy \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{2n^2}< \frac{1}{4}\) (Đpcm)

Answer 2

Answer 3

BƯỚC 1: THẤY ĐÚNG ĐÂU:

(....)=(...) KHÔNG HIỂU . bước này nhằm mục đích gì