Value of $\frac{1}{\sqrt{3}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{11}}+\frac{1}{\sqrt{11}+\sqrt{15}}+\cdots$ ($n$ terms)
Multiplying top and bottom by the denominator with a $-$ sign, we can rewrite this as
$$\frac{\sqrt 7 - \sqrt 3}{7 - 3} + \frac{\sqrt{11} - \sqrt 7}{11 - 7} + \frac{\sqrt{15} - \sqrt{11}}{15 - 11} + \dots + \text{ final term}$$
This does telescope.
Note that the $n^{th}$ term is $$\dfrac1{\sqrt{4n-1} + \sqrt{4n+3}}$$ Rationalize the denominator to get $$\dfrac1{\sqrt{4n-1} + \sqrt{4n+3}} = \dfrac{\sqrt{4n+3} - \sqrt{4n-1}}4$$ Now use telescopic cancelation to obtain what you want.