Question

10. CMR:

\(\sqrt{\text{1+2+3+...+(n−1)+n+(n−1)+...+3+2+1 }}\) = n

Answer 1

\(\sqrt{1+2+3+...+\left(n-1\right)+n+\left(n-1\right)+...+3+2+1}\)

\(=\sqrt{2\left(1+2+3+...+\left(n-1\right)\right)+n}\)

\(=\sqrt{2\cdot\left(\dfrac{\left(n-1\right)\left(n-1+1\right)}{2}\right)+n}\)

\(=\sqrt{n\left(n-1\right)+n}=\sqrt{n^2}=n\)