Estonian Math Competitions

We will denote $$s=\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\cdots +\sqrt{1+\frac{1}{2021^2}+\frac{1}{2022^2}}$$ Notice that $$1+\frac{1}{k^2}+\frac{1}{(k+1{)}^2}=\frac{k^2(k+1{)}^2+(k+1{)}^2+k^2}{k^2(k+1{)}^2}=\frac{(k(k+1)+1{)}^2}{(k(k+1){)}^2},$$ which yields $$\sqrt{1+\frac{1}{k^2}+\frac{1}{(k+1{)}^2}}=\frac{k(k+1)+1}{k(k+1)}=1+\frac{1}{k(k+1)}=1+\frac{1}{k}-\frac{1}{k+1}.$$ Therefore $$ s = 2021 + \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \dots + \left(\frac{1}{2021} - \frac{1}{2022}\right) = 2021 + \frac{2021}{2022}.