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) Câu trả lời: 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

Trần Hoàng Bảo Ngọc

18 tháng 10 2016

Help me

#Toán lớp 7

Trần Minh Hưng

1 tháng 11 2016

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)

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