Question

[Lê Nam Khánh0103]

cho A=\(\frac{36}{1.3.5}+\frac{36}{2.5.7}+\frac{36}{5.7.9}+.....+\frac{36}{25.27.29}\)        chứng minh A<3

Answer 1

[Nguyễn Linh Chi]

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

\(=9\left(\frac{4}{3.\left(1.5\right)}+\frac{4}{5\left(3.7\right)}+\frac{4}{7.\left(5.9\right)}+.....+\frac{4}{27\left(25.29\right)}\right)\)

\(=9\left(\frac{1}{3}\left(1-\frac{1}{5}\right)+\frac{1}{5}\left(\frac{1}{3}-\frac{1}{7}\right)+\frac{1}{7}\left(\frac{1}{5}-\frac{1}{9}\right)+...+\frac{1}{27}\left(\frac{1}{25}-\frac{1}{29}\right)\right)\)

\(=9\left(\frac{1}{3}-\frac{1}{3.5}+\frac{1}{5.3}-\frac{1}{5.7}+\frac{1}{7.5}-\frac{1}{7.9}+...+\frac{1}{27.25}-\frac{1}{27.29}\right)\)

\(=9\left(\frac{1}{3}-\frac{1}{27.29}\right)< 9.\frac{1}{3}=3\)

Vậy A < 3.