If
[tex]S=1+2+\cdots+(n-1)+n[/tex]
then we also have
[tex]S=n+(n-1)+\cdots+2+1[/tex]
Pairing up the terms, we can find that
[tex]2S=(1+n)+(2+n-1)+\cdots+(n-1+2)+(n+1)[/tex]
[tex]2S=(n+1)+(n+1)+\cdots+(n+1)+(n+1)[/tex]
[tex]S[/tex] consists of [tex]n[/tex] terms, so there are [tex]n[/tex] copies of [tex]n+1[/tex] in the sum [tex]2S[/tex]. Then we find an immediate closed form for the sum:
[tex]2S=n(n+1)\implies S=\dfrac{n(n+1)}2[/tex]
