Question

So sánh A = \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{98^2}+\dfrac{1}{99^2}\) và B = \(\dfrac{304}{1975}\)

Answer 1

Ta có:

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

\(A>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)

\(A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

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

\(A>\dfrac{49}{100}\)

Ta lại có:

\(\dfrac{49}{100}=\dfrac{96775}{197500}\)

\(\dfrac{304}{1975}=\dfrac{30400}{197500}\)

\(\Rightarrow\dfrac{49}{100}>\dfrac{304}{1975}\)

Mà \(A>\dfrac{49}{100}\)

\(\Rightarrow A>B\)