$\begin{align}  & \sum\nolimits_{r=1}^{n}{r}=\tfrac{n(n+1)}{2} \\ & 1+2+3+\cdot \cdot \cdot +n=\frac{n(n+1)}{2} \\ & \text{When }n=1 \\ & L.H.S=1,\text{ }R.H.S=\frac{1(1+1)}{2}=\frac{2}{2}=1 \\ & L.H.S=R.H.S \\ & \text{when }n=2 \\ & L.H.S \\ & 1+2=3 \\ & R.H.S=\frac{2(2+1)}{2}=\frac{2\times 3}{2}=3 \\ & L.H.S=R.H.S \\ & \text{The formula hold true for }n=1,2 \\ & \text{let }n=k \\ & 1+2+3+\cdot \cdot \cdot +k=\frac{k(k+1)}{2} \\ & \text{To test whether it hold true for }k+1\text{ }term \\ & \underbrace{1+2+3+\cdot \cdot \cdot +k}_{\tfrac{k(k+1)}{2}}+(k+1)=\frac{k(k+1)}{2}+(k+1) \\ & \underbrace{1+2+3+\cdot \cdot \cdot +k}_{\tfrac{k(k+1)}{2}}+(k+1)=\frac{k(k+1)+k(k+1)}{2}=\frac{{{k}^{2}}+k+2k+2}{2} \\ & \underbrace{1+2+3+\cdot \cdot \cdot +k}_{\tfrac{k(k+1)}{2}}+(k+1)=\frac{({{k}^{2}}+3k+2)}{2}=\frac{(k+1)(k+2)}{2} \\ & \text{The preposition is also valid for }k+1\text{ terms} \\\end{align}$