$\begin{align}  & \sum\nolimits_{r=1}^{n}{{{r}^{2}}}=\tfrac{n}{6}(n+1)(2n+1) \\ & {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+\cdot \cdot \cdot +{{n}^{2}}=\frac{n}{6}(n+1)(2n+1) \\ & \text{When }n=1 \\ & L.H.S={{1}^{2}} \\ & R.H.S=\frac{1}{6}(1+1)(2+1)=1 \\ & \text{The preposition is valid for }n \\ & \text{Assume that }n=k \\ & {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+\cdot \cdot \cdot +{{k}^{2}}=\frac{k}{6}(k+1)(2k+1) \\ & \text{To test whether the preposition hold true for (}k+1)\text{ terms} \\ & \underbrace{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+\cdot \cdot \cdot +{{k}^{2}}}_{\frac{k}{6}(k+1)(2k+1)}+{{(k+1)}^{2}} \\ & \frac{k}{6}(k+1)(2k+1)+{{(k+1)}^{2}}=(k+1)\left[ \frac{k(2k+1)+6(k+1)}{6} \right] \\ & \frac{k}{6}(k+1)(2k+1)+{{(k+1)}^{2}}=(k+1)\left[ \frac{2{{k}^{2}}+k+6k+6}{6} \right] \\ & \frac{k}{6}(k+1)(2k+1)+{{(k+1)}^{2}}=\frac{(k+1)(2{{k}^{2}}+7k+6)}{6} \\ & \frac{k}{6}(k+1)(2k+1)+{{(k+1)}^{2}}=\frac{(k+1)(k+2)(2k+3)}{6} \\ & \frac{k}{6}(k+1)(2k+1)+{{(k+1)}^{2}}=\frac{(k+1)(k+1+1)[2(k+1)+1]}{6} \\ & \text{The preposition hold true for }k+1,\text{ the formular is valid} \\\end{align}$