Question

Chứng tỏ rằng:

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{49^2}+\frac{1}{50^2}<1\)

Answer 1

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{49^2}+\frac{1}{50^2}\) 

< \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{48.49}+\frac{1}{49.50}\)

< \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{48.49}+\frac{1}{49.50}=1-\frac{1}{50}

Answer 2

Ta có:

\(\frac{1}{2^2}

Answer 3

ta có :

\(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2}\)

\(\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3}\)

\(\frac{1}{4^2}=\frac{1}{4.4}< \frac{1}{3.4}\)

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

\(\frac{1}{49^2}=\frac{1}{49.49}< \frac{1}{48.49}\)

\(\frac{1}{50^2}=\frac{1}{50.50}< \frac{1}{49.50}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+......+\frac{1}{49^2}+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{48.49}+\frac{1}{49.50}\)

ta có : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{48.49}+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{48}-\frac{1}{49}+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}\)

\(=\frac{49}{50}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+........+\frac{1}{49^2}+\frac{1}{50^2}< \frac{49}{50}\) ( 1 )

mà \(\frac{49}{50}< 1\) ( 2 )

từ ( 1 ) và ( 2 ) \(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+........+\frac{1}{49^2}+\frac{1}{50^2}< 1\)

\(\Rightarrow\text{Đ}PCM\)