Question

[Thảo Mai Phù Thủy]

CMR:\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{199}{99^2.100^2}\)<1

Answer 1

[Minh Triều]

\(\text{ta thấy }\frac{3}{1^2.2^2}=\frac{1}{1^2}-\frac{1}{2^2};\frac{5}{2^2.3^2}=\frac{1}{2^2}-\frac{1}{3^2};....;\frac{199}{99^2.100^2}=\frac{1}{99^2}-\frac{1}{100^2}\)

\(\text{suy ra }\)\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{199}{99^2.100^2}=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{99^2}-\frac{1}{100^2}\)

\(=\frac{1}{1^2}-\frac{1}{100^2}=\frac{1}{1}-\frac{1}{10000}=\frac{10000}{10000}-\frac{1}{10000}=\frac{9999}{10000}

Answer 2

[bímậtnhé]

\(\Rightarrow\frac{3}{1.4}+\frac{5}{4.9}+...+\frac{199}{9801.1000}\)

\(\Rightarrow1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{9801}-\frac{1}{1000}\)

\(\Rightarrow1-\frac{1}{1000}< 1\)