Question

If $S_n={\sum }_{r=1}^{n}tr=\frac{1}{6}n(2n^2+9n+13)$, then find ${\sum }_{r=1}^n\sqrt{tr}$

Answer 1

$S_n=\sum _{r=1}^n{t}_r=\frac{1}{6}n(2n^2+9n+13)$

$\therefore$ Putting $n=1$,

We get

$t_1=\frac{1}{6}(2+9+13)$

$\therefore t_1=4$ ……..$\left(1\right)$

$\therefore t_2+t_1=\frac{2}{6}(8+18+13)$

$\therefore t_2+t_1=13$

$\therefore t_2=9$ ……..$\left(2\right)$

$\therefore t_2+t_1+t_3=\frac{3}{6}(18+27+13)$

$\therefore t_3=16$ ………$\left(3\right)$

From $\left(1\right),\left(2\right)$ & $\left(3\right)$

$\therefore t_n=(n+1)$

$\therefore \sum _{r=1}^n\sqrt{t_r}=\sum _{r=1}^n\sqrt{(n+1{)}^2}$

$=\sum _{r=1}^n(n+1)$

$=\sum _{r=1}^{n}n+\sum _{r=1}^{r=n}1$

$=\frac{n(n+1)}{2}+n$.