If n > 1, what is the smallest positive integer n such that the expression \(\sqrt{1+2+3+...+n}\) simplifies to an integer?

\(\sqrt{1+2+3+...+n}\)=\(\sqrt{6+...+n}\)

Because n is the smallest and n > 1 so n = 2

\(\Rightarrow\sqrt{6+2}\)=\(\sqrt{8}\)=\(\sqrt[\sqrt{2}]{2}\) (elminate)

=>n=3

=>\(\sqrt{6+3}\)=\(\sqrt{9}=3\)(possible)

So n=3