Question

Bài 3 : CMR :1/4^2 + 1/6^2 + 1/8^2 + ... + 1/(2n)^2 < 1/4 ( n lớn hơn hoặc = 2)

Answer 1

Ta có

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

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

\(=\frac{1}{4}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\right)\)

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

\(=\frac{1}{4}.\left(1-\frac{1}{n}\right)< \frac{1}{4}.1=\frac{1}{4}\)

=> ĐPCM