Question

chứng tỏ rằng : A=\(\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+....+\frac{36}{25.27.29}< 3\)

Answer 1

Ta có:

\(A=\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}\)

\(\Rightarrow A=9.\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\right)\)

\(\Rightarrow A=9.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)

\(\Rightarrow A=9.\left(\frac{1}{1.3}-\frac{1}{27.29}\right)\)

\(\Rightarrow A=9.\left(\frac{1}{3}-\frac{1}{783}\right)\)

\(\Rightarrow A=9.\frac{1}{3}-9.\frac{1}{783}\)

\(\Rightarrow A=3-\frac{1}{87}\)

Vì \(3-\frac{1}{87}< 3.\)

\(\Rightarrow A< 3\left(đpcm\right).\)

Chúc bạn học tốt!