Question

${\sum }_{r=1}^{n}si{n}^{-1}\left(\frac{\sqrt{r}-\sqrt{r-1}}{\sqrt{r(r+1)}}\right)$ is equal to

• A.
  
• B.
  
• C.
  
• D.
  

Answer 1

The correct option is C

$si{n}^{-1}\left(\frac{\sqrt{r}-\sqrt{r-1}}{\sqrt{r(r+1)}}\right)=ta{n}^{-1}\left(\frac{\sqrt{r}-\sqrt{r-1}}{\sqrt{r(r+1)}}\right)$
$=ta{n}^{-1}\sqrt{r}-ta{n}^{-1}(\sqrt{r}-1)$
$\Rightarrow {\sum }_{r=1}^{n}si{n}^{-1}\left(\frac{\sqrt{r}-\sqrt{r-1}}{\sqrt{r(r+1)}}\right)={\sum }_{r=1}^n(ta{n}^{-1}\sqrt{r}-ta{n}^{-1}(\sqrt{r-1}\left)\right)$
$=ta{n}^{-1}\sqrt{n}$ (By method of Differences)