Question

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

• A. $\frac{1}{2}n\left(n+1\right)$
• B. $\frac{1}{2}n\left(n+3\right)$
• C. ${\left(n+1\right)}^2$
• D. $n^2$

Answer 1

Answer: (B)

$t_r=S_r-S_{r-1}=\frac{1}{6}r\left(2r^2+9r+13\right)$
$-\frac{1}{6}(r-1)\left\{2(r-1{)}^2+9(r-1)+13\right\}$
$=\frac{2}{6}\left(r^3-(r-1{)}^3\right)+\frac{9}{6}\left(r^2-(r-1{)}^2\right)+\frac{13}{6}(r-(r-1))$
$=\frac{1}{3}\left(3r^2-3r+1\right)+\frac{3}{2}(2r-1)+\frac{13}{6}$
$=r^2+2r+1=(r+1{)}^2$
$\Rightarrow \sqrt{t_r}=(r+1)$
$\Rightarrow {\sum }_{r=1}^n\sqrt{t_r}={\sum }_{r=1}^n(r+1)$
$={\sum }_{r=1}^{n}r+{\sum }_{r=1}^{n}1=\frac{n(n+1)}{2}+n=\frac{n(n+3)}{2}$