Đặt \(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right).........\left(\frac{1}{100^2}-1\right)\)
\(\Rightarrow A=\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}.\frac{1-4^2}{4^2}............\frac{1-100^2}{100}\)
\(\Rightarrow A=\frac{-3}{2^2}.\frac{-8}{3^2}.\frac{-15}{4^2}............\frac{-9999}{100^2}\)
\(\Rightarrow A=\frac{-1.3}{2^2}.\frac{-2.4}{3^2}.\frac{-3.5}{4^2}...............\frac{-99.101}{100^2}\)
\(\Rightarrow A=\frac{-\left(1.2.3.............99\right).\left(3.4.5............101\right)}{\left(2.3.4......100\right).\left(2.3.4.............100\right)}\)
\(\Rightarrow A=\frac{-1.101}{100.2}=\frac{-101}{200}\)
Vậy \(A=\frac{-101}{200}\)
Chúc bn học tốt
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)
=>\(\)
=>\(A< -\left(\frac{1.2.3....99}{2.3.4...100}\right)=-\frac{1}{100}\)
Mà \(-\frac{1}{100}>-\frac{1}{2}\)
=>\(A>-\frac{1}{2}\) đúng ko nhỉ
\(\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right).....\left(\frac{1}{100^2}-1\right)\)
= \(\frac{3}{2.2}.\frac{8}{3.3}.\frac{15}{4.4}.....\frac{9999}{100.100}\)
= \(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{99.101}{100.100}\)
=\(\frac{1.2.3.4.5.....99}{2.3.4.5.....100}.\frac{3.4.5.6.7.....101}{2.3.4.5.6.....100}\)
=\(\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)
ta có \(\frac{-1}{2}< 0< \frac{101}{200}\Rightarrow\frac{101}{200}>\frac{-1}{2}\)
