Question

Với n thuộc N*

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

 

Answer 1

Ta có: \(1+2+3+...+\left(n+1\right)=\frac{\left(n+1\right)\left(n+2\right)}{2}=\frac{n^2+3n+2}{2}\)

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

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

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

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