Đặt \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
Ta có:
\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
\(A=\frac{1}{2^2}\cdot\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
Đặt \(B=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
\(\Rightarrow A=\frac{1}{2^2}\cdot B=\frac{1}{4}\cdot B\)
Ta thấy:
\(\frac{1}{1^2}=\frac{1}{1^2};\frac{1}{2^2}< \frac{1}{1\cdot2};\frac{1}{3^2}< \frac{1}{2\cdot3};...;\frac{1}{50^2}< \frac{1}{49\cdot50}\)
\(\Rightarrow\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{1}{1^2}+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}\)
hay \(B< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow B< 2+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{49}-\frac{1}{49}\right)-\frac{1}{50}\)
\(\Rightarrow B< 2-\frac{1}{50}\)
\(\Rightarrow\frac{1}{4}\cdot B< \frac{1}{4}\cdot\left(2-\frac{1}{50}\right)\)
hay \(A< \frac{1}{2}-\frac{1}{200}\) (1)
Vì \(\frac{1}{200}>0\)
\(\Rightarrow\frac{1}{2}-\frac{1}{200}< \frac{1}{2}\) (2)
Từ (1) và (2)
\(\Rightarrow A< \frac{1}{2}\) (đpcm)
help me
