Question

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

Answer 1

Dễ quá

Answer 2

ohh

Answer 3

Giúp mình

Answer 4

A=1/2²+1/3²+1/4²+...+1/98²+1/99²

Ta thấy:

1/2²=1/2.2 < 1/1.2

1/3²=1/3.3 < 1/2.3

1/4²=1/4.4 < 1/3.4

...

1/100²=1/100.100 < 1/99.100

A<1/1.2+1/2.3+1/3.4+...+1/99.100

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

A<1/1-1/100

A<99/100

Mk mới làm đc tới đây thôi!

Answer 5

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)

\(A< \dfrac{1}{2^2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

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

\(A< \dfrac{1}{2^2}+\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{3}{4}-\dfrac{1}{100}< \dfrac{3}{4}\) (đpcm)