Question

Cho A = \(\frac{1}{1.3.5}\) + \(\frac{1}{3.5.7}\) + ..... + \(\frac{1}{47.49.51}\). Chứng minh A < \(\frac{1}{12}\)

Answer 1

Bài làm:

Ta có: \(A=\frac{1}{1.3.5}+\frac{1}{3.5.7}+...+\frac{1}{47.49.51}\)

\(A=\frac{1}{4}\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+...+\frac{4}{47.49.51}\right)\)

\(A=\frac{1}{4}\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{47.49}-\frac{1}{49.51}\right)\)

\(A=\frac{1}{4}\left(\frac{1}{3}-\frac{1}{49.51}\right)\)

\(A=\frac{1}{12}-\frac{1}{4.49.51}< \frac{1}{12}\)

Vậy \(A< \frac{1}{12}\)

Answer 2

Từ đề bài suy ra\(4A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+...+\frac{4}{47.49.51}\)

\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{47.49}-\frac{1}{49.51}=\frac{1}{3}-\frac{1}{49.51}< \frac{1}{3}\)

\(\Rightarrow A< \frac{1}{12}\left(đpcm\right)\)

Answer 3

\(4A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+...+\frac{4}{47.49.51}\)

\(4A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{47.49}-\frac{1}{49.51}\)

\(4A=\frac{1}{1.3}-\frac{1}{49.51}=\frac{1}{3}-\frac{1}{2499}< \frac{1}{3}\)

=>    \(A< \frac{1}{3}:4=\frac{1}{12}\)

VẬY TA CÓ ĐPCM.

NẾU ĐÚNG MONG MỌI NGƯỜI ỦNG HỘ Ạ

Answer 4

\(A=\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+...+\frac{1}{47\cdot49\cdot51}\)

\(A=\frac{1}{4}\left(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+...+\frac{4}{47\cdot49\cdot51}\right)\)

\(A=\frac{1}{4}\left(\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{47\cdot49}-\frac{1}{49\cdot51}\right)\)

\(A=\frac{1}{4}\left(\frac{1}{3}-\frac{1}{49\cdot51}\right)=\frac{208}{2499}\)

Đến đây tự chứng minh...

Answer 5

\(A=\frac{1}{1.3.5}+\frac{1}{3.5.7}+...+\frac{1}{47.49.51}\)

\(A=\frac{1}{4}\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+...+\frac{4}{47.49.51}\right)\)

\(A=\frac{1}{4}\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{47.49}+\frac{1}{49.51}\right)\)

\(A=\frac{1}{4}\left(\frac{1}{3}-\frac{1}{49.51}\right)\)

\(A=\frac{1}{12}-\frac{1}{4.49.51}< 12\)

Vậy \(A< 12\)