## Precalculus (10th Edition)

$S=1+2+...+(n-1)+n$ $S=n+(n-1)+...+2+1$ Hence $2S=(1+n)+(2+(n-1)+...+((n-1)+2)+(n+1)=(n+1)+(n+1)+...+(n+1)+(n+1)\\S=\frac{n(n+1)}{2}.$ Because in $2S$ there are $n$ pieces of $n+1$ terms.
$S=1+2+...+(n-1)+n$ $S=n+(n-1)+...+2+1$ Hence $2S=(1+n)+(2+(n-1)+...+((n-1)+2)+(n+1)=(n+1)+(n+1)+...+(n+1)+(n+1)\\S=\frac{n(n+1)}{2}.$ Because in $2S$ there are $n$ pieces of $n+1$ terms.