Question

If $S_{n}=\displaystyle \sum _{r = 1}^{n} t_{r} = \frac{1}{6} n\left(2 n^{2} + 9 n + 13\right),$ then $\displaystyle \sum _{r = 1}^{n} \sqrt{t_{r}}$ 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)

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