Question

 cho A=\(\frac{1}{1^2}+\frac{1}{2^2}+......+\frac{1}{50^2}\)Chưng minh rang a<2

Answer 1

A=1+[\(\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{50^2}\)

ta có \(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};......;\frac{1}{50^2}<\frac{1}{49.50}\)

=>A<1+\(\left[\frac{1}{1.2}+.........+\frac{1}{49.50}\right]\)

=>A<1+\(\left[\frac{1}{1}-\frac{1}{50}\right]\)

=>A<1+\(\frac{49}{50}\)

=>A<\(\frac{99}{50}\) <2

=>A<2

K MÌNH NHA BÀI NÀY MÌNH GHI MỎI TAY LẮM

Answer 2

A=\(\frac{1}{1^2}+\frac{1}{2^2}+....+\frac{1}{50^2}\)

A<\(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49\cdot50}\)

A<1+\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)\)

A<1+\(\left(1-\frac{1}{50}\right)\)

A<1+\(\frac{49}{50}\)

=>A<2