$$\frac{1}{\sqrt a_1 + \sqrt a_2} + \frac{1}{\sqrt a_2 + \sqrt a_3} + \cdots + \frac{1}{\sqrt a_{n-1} + \sqrt a_n} = \frac{n-1}{\sqrt a_1 + \sqrt a_n}$$
I need to prove this; I don't know how to get from left side terms to the right side term? What should I do to figure this out?
Please answer with detail as much as possible, thanks!
Just rationalise the terms. The denominator becomes same for all, $d$, that is the common difference of AP:
$$S = \sum_{k = 1}^{n-1}\frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{d} = \frac{\sqrt{a_{n}}-\sqrt{a_1}}{d} = \frac{n-1}{\sqrt{a_{n}}+\sqrt{a_1}}$$