Question

Bài 21: Chứng tỏ rằng: ` 1/ 2^2 + 1/3^2 + 1/4^2 + ... + 1/100^2 < 3/4 `

Answer 1

Làm lại ạ: \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
\(A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{100.100}\)
\(A< \dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{100.101}\)
\(A< \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{100}-\dfrac{1}{101}\)
\(A< \dfrac{1}{2}-\dfrac{1}{101}\)
\(A< \dfrac{101}{202}-\dfrac{2}{202}\)
\(A< \dfrac{99}{202}< \dfrac{3}{4}\)
\(\Rightarrow A< \dfrac{3}{4}\)

Answer 2

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
\(A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{100.100}\)
\(A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A< 1-\dfrac{1}{100}\)
\(A< \dfrac{100}{100}-\dfrac{1}{100}\)
\(A< \dfrac{99}{100}< \dfrac{3}{4}\)
\(\Rightarrow A< \dfrac{3}{4}\)