Question

Chứng minh rằng:

1+1/2₂+1/3₂+1/4₂+...+1/100₂<2

giúp mk vs nha!!!!!!!!!!!!!!!!!!!!!!

Answer 1

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

Ta có:

\(\dfrac{1}{2^2}< \dfrac{1}{2.3}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)

...

\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

Mà

\(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

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

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

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

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

Vậy \(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)<2

Answer 2

Ta có: \(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}=1+\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1+\left(1-\dfrac{1}{100}\right)< 2\left(đpcm\right)\)

Answer 3

ta có:Gọi biểu thức trên là A

Ta có:

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

A<1+\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+........+\dfrac{1}{99.100}\)

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

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

=)A<2

Answer 4

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

Ta thấy:

\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3\cdot4}\)

\(...............\)

\(\dfrac{1}{100^2}< \dfrac{1}{99\cdot100}\)

\(\Rightarrow1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)

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

\(\Rightarrow A< 1+\left(1-\dfrac{1}{100}\right)\)

\(\Rightarrow A< 1+\left(\dfrac{100}{100}-\dfrac{1}{100}\right)\)

\(\Rightarrow A< 1+\dfrac{99}{100}\)

Mà \(\dfrac{99}{100}< 1\Rightarrow1+\dfrac{99}{100}< 2\)

Vậy \(A< 2\).

Answer 5

Chứng minh rằng : 1/22 + 1/32 + 1/42 + ... + 1/1002 < 1.. - Miny.vn - Cộng đồng hỗ trợ học tập