Đặt \(A=\sqrt{1+2+3+4+...+\left(n-1\right)+n+\left(n-1\right)+...+3+2+1}\)
\(A=\sqrt{2\left(1+2+...+n-1\right)+n}\)
\(A=\sqrt{\frac{2\left(n-1\right)n}{2}+n}=\sqrt{n^2}=n\)
Vậy: \(\sqrt{1+2+3+4+...+\left(n-1\right)+n+\left(n-1\right)+...+3+2+1}\)=n