\(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\sqrt{\frac{n^2\left(n+1\right)^2+n^2+\left(n+1\right)^2}{n^2\left(n+1\right)^2}}=\sqrt{\frac{\left[n\left(n+1\right)\right]^2+2n^2+2n+1}{n^2\left(n+1\right)^2}}\)
\(=\sqrt{\frac{\left[n\left(n+1\right)\right]^2+2n\left(n+1\right)+1}{n^2\left(n+1\right)^2}}=\sqrt{\frac{\left[n\left(n+1\right)+1\right]^2}{n^2\left(n+1\right)^2}}=\frac{n\left(n+1\right)+1}{n\left(n+1\right)}=1+\frac{1}{n\left(n+1\right)}\)
Vậy:
\(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}}\)
\(=1+\frac{1}{1.2}+1+\frac{1}{2.3}+...+1+\frac{1}{99.100}\)
\(=99+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=99+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=99+1-\frac{1}{100}=100-\frac{1}{100}=...\)
