Suppose the value of the sum is [tex]S[/tex]:
[tex]S=\displaystyle\sum_{k=1}^nk[/tex]
So
[tex]S=1+2+3+\cdots+(n-2)+(n-1)+n[/tex]
but also
[tex]S=n+(n-1)+(n-2)+\cdots+3+2+1[/tex]
That is,
[tex]S=\displaystyle\sum_{k=1}^n(n-k+1)[/tex]
Adding these together, we have
[tex]2S=\displaystyle\sum_{k=1}^n(k+n-k+1)=\sum_{k=1}^n(n+1)=(n+1)\sum_{k=1}^n1=n(n+1)[/tex]
[tex]\implies S=\dfrac{n(n+1)}2[/tex]
