Cool Summation

$\sum_{n=2}^{r} \frac{1}{\sqrt{(2n+7)}+\sqrt{(2n+5)}}=1$ On rationalizing we get, $\sum_{n=2}^{r} \frac{\sqrt{(2n+7)}-\sqrt{(2n+5)}}{2n +7 -(2n+5)}=1$ $\Rightarrow \sum_{n=2}^{r} \sqrt{(2n+7)}-\sqrt{(2n+5)}=2$ $\Rightarrow \sqrt{(2r+7)} -\sqrt{(2r+5)}+\sqrt{(2(r-1)+7)}$ $-\sqrt{(2(r-1)+5)} +.... +\sqrt{2\times 2 +7}-\sqrt{(2\times 2+5)}=2$
All the consecutive terms get cancelled leaving the first and the last term on the $LHS$ .

$\Rightarrow \sqrt{(2r+7)}-\sqrt{(9)}=2$ $\Rightarrow \sqrt{(2r+7)}=5$ On solving we get, $\boxed{r=9}$

$\therefore$ The value of $r$ is $9$ .

Step1: Rationalise the denominator as follows: $\frac{1}{\sqrt{2n+7} +\sqrt{2n+5}} \times\frac{\sqrt{2n+7} - \sqrt{2n+5}}{\sqrt{2n+7} - \sqrt{2n+5}}$ $= \frac{\sqrt{2n+7} - \sqrt{2n+5}}{2}$ Step 2: calculate the telescoping sum by cancellation: $\frac{1}{2} \displaystyle\sum_{n=2}^{r} (\sqrt{2n+7} - \sqrt{2n+5}) = (\sqrt{11} - \sqrt{9}) + (\sqrt{13} - \sqrt{11}) +...$ $+(\sqrt{2r+5} - \sqrt{2n+3}) +(\sqrt{2r+7} - \sqrt{2r+5}) = \frac{1}{2} (\sqrt{2r+7} - 3)$ Step 3: Equate to 1: $\frac{1}{2} (\sqrt{2r+7} - 3) = 1 \Rightarrow\sqrt{2r+7} = 5 \Rightarrow\ 2r+7 = 25$ $\therefore\boxed{r=9}$

Though similar idea with Tejas's, I believe this is simpler.

$\displaystyle \sum _{n=2} ^r {\dfrac {1} {\sqrt{2n+7}+\sqrt{2n+5}} } = 1$

$\Rightarrow \displaystyle \sum _{n=2} ^r {\dfrac {\sqrt{2n+7}-\sqrt{2n+5}} {2n+7-2n-5}} = 1\quad \Rightarrow \sum _{n=2} ^r {\left( \sqrt{2n+7}-\sqrt{2n+5} \right) } = 2$

$= \sqrt{11}-\sqrt{9} + \sqrt{13}-\sqrt{11} + \sqrt{15}-\sqrt{13} + ... + \sqrt{2r+7}-\sqrt{2r+5}$

$= -\sqrt{9} + \sqrt{2r+7} = 2\quad \Rightarrow \sqrt{2r+7} = 5 \quad \Rightarrow 2r+7 = 25$

$\Rightarrow 2r = 18 \quad \Rightarrow r = \boxed{9}$