Ta có: \(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}}\)
\(=1+1-\frac{1}{1+1}+1+\frac{1}{2}-\frac{1}{2+3}+...+1+\frac{1}{99}-\frac{1}{1+99}\)
\(=100-\frac{1}{100}=\frac{9999}{100}\)