Question

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

Answer 1

Ta có:

A=\(1+2+3+...+\left(n-1\right)+n+\left(n-1\right)+...+3+2+1\)

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

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

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

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

Vậy \(\sqrt{A}=\sqrt{n^2}=n\)

Answer 2

Ta có :

A = 1 + 2 + 3 + ... + ( n - 1 ) + n + ( n - 1 ) + ... + 3 + 2 + 1

   = 2 ( 1 + 2 + 3 + ... + ( n - 1 ) + n

   = 2 ( n . ( n - 1 ) /2 ) + n

   = n ( n - 1 ) + n = n ( n - 1 + 1 ) = n²

Vậy \(\sqrt{A}=\sqrt{n^2}=n\)