Question

[vvvvvvvv]

Chứng minh rằng: .

câu 5 có phải \(\dfrac{25}{147}\) không vậy mọi người

Answer 1

[Murana Karigara]

1)\(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}\)

\(A< \dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{99.100}\)

\(A< \dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A< \dfrac{1}{4}-\dfrac{1}{100}\)

\(A< \dfrac{1}{4}\)(1)

\(A>\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{100.101}\)

\(A>\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{100}-\dfrac{1}{101}\)

\(A>\dfrac{1}{5}-\dfrac{1}{101}\)

\(A>\dfrac{96}{505}>\dfrac{1}{6}\)

\(A>\dfrac{1}{6}\)(2)

Từ (1) và (2)

\(\dfrac{1}{6}< A< \dfrac{1}{4}\)

2)

\(A=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}+\dfrac{1}{95.98}\)

\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{98}\right)\)\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{98}\right)\)

\(A=\dfrac{1}{3}.\dfrac{24}{49}=\dfrac{8}{49}\)