Answer
True
Work Step by Step
Let $S=1+2+3+...+(n-1)+n$
By commutativity of addition, we can write
$S=n+(n-1) +(n-2)+ ...+ 2+1$
Adding term by term both the equations, we get
$2S=[1+n] + [2+(n-1)] +[3+(n-2)]+ ...+ [(n-1)+2]+[n+1]$
or simply, $2S=[1+n] + [1+n] +[1+n]+ ...+[1+n]+[1+n]$ (n times)
Thus, $2S=(n+1)\times n$
or $S=\frac{n(n+1)}{2}$
