Đề có cho n >=0 ko bạn?
\(\sqrt{1+2+3+....+\left(n-1\right)+n+\left(n-1\right)+...+3+2+1}\)
\(=\sqrt{2.\left[1+2+3+...+\left(n-1\right)\right]+n}=\sqrt{2.\frac{\left[\left(n-1\right)+1\right]\left(n-1\right)}{2}+n}\)
\(=\sqrt{\left(n-1+1\right)\left(n-1\right)+n}=\sqrt{n.\left(n-1\right)+n}=\sqrt{n^2-n+n}=n\)