Question

Cho A = \(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\). Chứng minh rằng A < \(\dfrac{7}{4}\)

Answer 1

Ta có:

A=\(1+\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{100.100}\)

A<\(1+\dfrac{1}{2.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

A<\(1+\dfrac{1}{4}+\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

A<\(\dfrac{5}{4}\)+\(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{99}-\dfrac{1}{100}\)

A<\(\dfrac{5}{4}+\dfrac{1}{2}-\dfrac{1}{100}\)

A<\(\dfrac{5}{4}+\dfrac{49}{100}\)

A<\(\dfrac{174}{100}\)<\(\dfrac{7}{4}\)

=>A<\(\dfrac{7}{4}\)

Tick giùm mink nha :D

Answer 2

1/2^2<1/2.3,1/3^2<1/2.3,.....,1/100^2<1/99.100

A<1+1/2.3+1/3.4+....+1/99.100

A<1+1/2-1/3+1/3-1/4+1/4-1/5+....+1/99-1/100

A<1+1/2-1/100

A<3/2-1/100 mà 3/2=6/4

A<6/4-1/100<7/4

A<7/4