\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{99^2}\right)\left(1-\dfrac{1}{100^2}\right)\)
\(=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)...\left(1-\dfrac{1}{99}\right)\left(1+\dfrac{1}{99}\right)\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{100}\right)\)
\(=\dfrac{1}{2}.\dfrac{3}{2}.\dfrac{2}{3}.\dfrac{4}{3}...\dfrac{98}{99}.\dfrac{100}{99}.\dfrac{99}{100}.\dfrac{101}{100}\)
\(=\dfrac{1.2...98.99}{2.3...99.100}.\dfrac{3.4...99.101}{2.3...99.100}\)
\(=\dfrac{1}{100}.\dfrac{101}{2}\)
\(=\dfrac{101}{200}\)
Bài này chỉ cần áp dụng HĐT số 3 là ra thôi nha bạn:
Đặt:
\(A=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{99^2}\right)\left(1-\dfrac{1}{100^2}\right)\)
Xét:
\(1-\dfrac{1}{2^2}=1^2-\dfrac{1}{2^2}=\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{2}\right)\)
\(1-\dfrac{1}{3^2}=1^2-\dfrac{1}{3^2}=\left(1+\dfrac{1}{3}\right)\left(1-\dfrac{1}{3}\right)\)
Tương tự ta có:
\(1-\dfrac{1}{100^2}=1^2-\dfrac{1}{100^2}=\left(1+\dfrac{1}{100}\right)\left(1-\dfrac{1}{100}\right)\)
\(\Rightarrow A=\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)\left(1-\dfrac{1}{4}\right)...\left(1+\dfrac{1}{99}\right)\left(1-\dfrac{1}{99}\right)\left(1+\dfrac{1}{100}\right)\left(1-\dfrac{1}{100}\right)\)
\(\Rightarrow A=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\left(1+\dfrac{1}{4}\right)...\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{100}\right)\)
\(\Rightarrow A=\dfrac{1}{2}.\dfrac{3}{2}.\dfrac{2}{3}.\dfrac{4}{3}.\dfrac{3}{4}.\dfrac{5}{4}.....\dfrac{99}{100}.\dfrac{101}{100}\)
\(\Rightarrow A=\dfrac{1.3.2.4.3.5.....99.101}{2.2.3.3.4.4.....100.100}\)
\(\Rightarrow A=\dfrac{1.2.3....99}{2.3.4....100}.\dfrac{3.4.5.....101}{2.3.4......100}\)
\(\Rightarrow A=\dfrac{1}{100}.\dfrac{101}{2}=\dfrac{101}{200}\)
