Question

Chứng minh rằng : 

1/1² + 1/2² + 1/3² + 1/4² + ... + 1/50²   < 2

 

Answer 1

\(\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{50^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

\(\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{50^2}< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{50^2}< 1+1-\frac{1}{50}< 2\)

\(\Rightarrow\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{50^2}< 2\)

Answer 2

\(\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{50^2}=\frac{1}{1.1}+\frac{1}{2.2}+...+\frac{1}{50.50}\)

Mà \(\frac{1}{1.1}+\frac{1}{2.2}+...+\frac{1}{50.50}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

Lại có:

\(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1+1-\frac{1}{50}=1+\frac{49}{50}< 2\RightarrowĐpcm\)

Answer 3

1/1² + 1/2² + 1/3² + 1/4² + ... + 1/50²   < 1 + (\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+.......+\(\frac{1}{49.50}\))

1/1² + 1/2² + 1/3² + 1/4² + ... + 1/50² < 1+ ( 1- 1/2+1/2-1/3+1/3-.......+1/49-1/50)

1/1² + 1/2² + 1/3² + 1/4² + ... + 1/50²< 1+ ( 1- 1/50)

1/1² + 1/2² + 1/3² + 1/4² + ... + 1/50²< 1+ 49/50 

vì 1+49/50< 2 => 1/1² + 1/2² + 1/3² + 1/4² + ... + 1/50² <2

Answer 4

Đặt A = \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\)

Ta có :\(\frac{1}{1^2}=1\)

\(\frac{1}{2^2}< \frac{1}{1.2}\)

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

................................

\(\frac{1}{50^2}< \frac{1}{49.50}\)

\(\Rightarrow A< 1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)

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

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

\(\Leftrightarrow A=2-\frac{1}{50}\left(đpcm\right)\)