Ta có:1+2+3+..+(n-1)
=>số số hạng của tổng trên là:\(\frac{\left(n-1\right)-1}{1}\) +1=n-2+1=n-1
vậy:1+2+3+..+(n-1)=[(n-1)+1].(n-1):2=n(n-1):2
=>\(\sqrt{1+2+3+...+\left(n-1\right)+n+\left(n-1\right)+..+3+2+1}\)
\(\sqrt{n\left(n-1\right):2.2+n}\)
\(\sqrt{n\left(n-1\right)+n}\)
\(\sqrt{n.n-n+n}\)
\(\sqrt{\sqrt{n}}\)=n
vậy\(\sqrt{1+2+3+...+\left(n-1\right)+n+\left(n-1\right)+..+3+2+1}\)
=n(dpcm)
\(\left\{\begin{matrix}n\ge1\\A=\sqrt{1+2+..+\left(n-1\right)+n+\left(n-1\right)+...+2+1}\end{matrix}\right.\)
\(B=1+2+....\left(n-2\right)+\left(n-1\right)\)
\(C=\left(n-1\right)+\left(n-2\right)+...+2+1\)
B+C=\(\left[1+\left(n-1\right)\right]+\left[2+\left(n-2\right)\right]+...+\left[\left(n-2\right)+2\right]+\left[\left(n-1\right)+1\right]\)
\(B+C=n.\left(n-1\right)\)
\(A=\sqrt{B+n+C}=\sqrt{n\left(n-1\right)+n}=\sqrt{n^2-n+n}=\sqrt{n^2}=\left|n\right|\)
\(\left\{\begin{matrix}n\ge1\\A=n\end{matrix}\right.\) => dpcm