Question

cho \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

và \(B=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{200^2}\)

Khi đó \(\frac{A}{B}=?\)

Answer 1

 

\(B=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+..+\frac{1}{100^2}=\frac{1}{\left(2.2\right)^2}+\frac{1}{\left(2.3\right)^2}+\frac{1}{\left(2.4\right)^2}+..+\frac{1}{\left(2.100\right)^2}\)

\(B=\frac{1}{2^2.2^2}+\frac{1}{2^2.3^2}+\frac{1}{2^2.4^2}+..+\frac{1}{2^2.100^2}=\frac{1}{2^2}.\frac{1}{2^2}+\frac{1}{2^2}.\frac{1}{3^2}+\frac{1}{2^2}.\frac{1}{4^2}+..+\frac{1}{2^2}.\frac{1}{100^2}\)

\(=\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)=\frac{1}{4}.A\)

\(=>\frac{A}{B}=\frac{A}{\frac{A}{4}}=4\)