Lời giải:
Dễ thấy:
\(\dfrac{1}{1^2}=1\)
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
\(....\)
\(\dfrac{1}{50^2}=\dfrac{1}{50.50}< \dfrac{1}{49.50}\)
\(\Rightarrow A< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(\Rightarrow A< 1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow A< 1+1-\dfrac{1}{50}\)
\(\Rightarrow A< 2-\dfrac{1}{50}\)
\(\Rightarrow A< 2\)
\(\text{Ta có : }A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+.......+\dfrac{1}{50^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.........+\dfrac{1}{50.51}\)
\(< =>A< 1-\dfrac{1}{51}=\dfrac{50}{51}< 1< 2\left(\text{đ}pcm\right)\)
