Mathematics - Sequence & Series Question with Solution | TestHub

We have ${\mathrm{t}}_{\mathrm{n}}={\mathrm{S}}_{\mathrm{n}}-{\mathrm{S}}_{\mathrm{n}-1}\forall \mathrm{n}\ge 2$

$$\begin{array}{rlrlr}& \therefore t_n=\frac{1}{6}\left[2\left(n^3-(n-1{)}^3\right)+9\left(n^2-(n-1{)}^2\right)+13(n-n+1)\right]& =\frac{1}{6}\left[6n^2-6n+2+9(2n-1)+13\right]& =\frac{1}{6}\left(6n^2+12n+6\right)=(n+1{)}^2& \therefore \sum _{r=1}^n\sqrt{t_r}=\sum _{r=1}^n(r+1)=\frac{1}{2}(n+1)(n+2)-1=\frac{1}{2}n(n+3)\end{array}$$