Question

chứng minh rằng

3/1^2.2^2+5/2^2.3^2+7/3^2.4^2+.....+19/9^2.10^2<1

Answer 1

\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+...+\dfrac{19}{9^{10}.10^2}\)

\(=\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{9^{10}}-\dfrac{1}{10^2}\)

\(=1-\dfrac{1}{10^2}< 1\)

\(\Rightarrowđpcm\)