Ta đi chứng minh công thức tổng quát: \(f\left(n\right)=\frac{2n+1+\sqrt{n\left(n+1\right)}}{\sqrt{n}+\sqrt{n+1}}=\left(n+1\right)\sqrt{n+1}-n\sqrt{n}\)
Thật vậy: \(\left[\left(n+1\right)\sqrt{n+1}-n\sqrt{n}\right]\left(\sqrt{n}+\sqrt{n+1}\right)=\left(n+1\right)\sqrt{n\left(n+1\right)}-n^2+\left(n+1\right)^2-n\sqrt{n\left(n+1\right)}=2n+1+\sqrt{n\left(n+1\right)}\)Áp dụng, ta được: \(f\left(1\right)+f\left(2\right)+...+f\left(2020\right)=\left(2\sqrt{2}-1\sqrt{1}\right)+\left(3\sqrt{3}-2\sqrt{2}\right)+\left(4\sqrt{4}-3\sqrt{3}\right)+...+\left(2021\sqrt{2021}-2020\sqrt{2020}\right)=2021\sqrt{2021}-1\)
