If $S_n={\displaystyle \sum _{r=1}^n{t}_r=\frac{1}{6}n\left(2n^2+9n+13\right)}$ , then $S_n={\displaystyle \sum _{r=1}^n\sqrt{t_r}}$ , equals

$S_2=\frac{1}{6}\times 2\left(8+18+13\right)=13$

$\therefore t_2=S_2-S_1=9$

$S_3=\frac{1}{6}\times 3\left[18+27+13\right]=29$

$\therefore t_3=S_3-S_2=29-13=16=4^2$

$\therefore {\displaystyle \sum _{r=1}^n\sqrt{t_r}=\sqrt{t_1}+\sqrt{t_2}+\cdots +\sqrt{t_n}=2+3+\cdots +\left(n+1\right)=\frac{n}{2}\left(n+3\right)}$