Question

If ${\text{S}}_{\text{n}}\text{=}\sum _{\text{r = 1}}^{\text{n}}{T}_r\text{=}\frac{\text{1}}{\text{6}}\text{n}\left(\text{2}{\text{n}}^{\text{2}}\text{+ 9n + 13}\right)\text{,then}\text{find}\sum _{\text{r = 1}}^{\text{n}}\sqrt{\text{tr}}$

Answer 1

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

$T_r=S_r-S_{r-1}$

$=\frac{r}{6}[2r^2+9r+13)-\frac{(r-1)}{6}\left[2\right(r-1{)}^2+9(r-1)+13]$

$=\frac{1}{6}[2r^3+9r^2+13r-(r-1\left)\right[2r^2+2-4r+9r-9+13\left]\right]$

$=\frac{1}{6}[6r^2+12r+6]$

$=r^2+2r+1$

$=(r+1{)}^2$

$\Rightarrow \sqrt{T_r}=(r+1)$

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

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

$=\frac{n(n+1)}{2}+n=\frac{n^2+n+2n}{2}=\frac{n(n+3)}{2}$