<s>[INST] How to answer this kind of questions How can I answer this kind of questions?
I should prove the following
$$\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \frac{1}{\sqrt{a_3} + \sqrt{a_4}} + ... + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}} = \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}$$

Sorry for my poor English
 [/INST] $$\frac{\sqrt{a_1}-\sqrt{a_2}}{{a_1} + {a_2}}+\frac{\sqrt{a_2}-\sqrt{a_3}}{{a_2}-{a_3}}+....+\frac{\sqrt{a_{n-1}}-\sqrt{a_n}}{a_{n-1}-{a_n}}=\frac{\sqrt{a_1}-\sqrt{a_2}}{-d}+\frac{\sqrt{a_2}-\sqrt{a_3}}{-d}+....+\frac{\sqrt{a_{n-1}}-\sqrt{a_n}}{-d}=\frac{\sqrt{a_1}-\sqrt{a_2}+\sqrt{a_2}-\sqrt{a_3}+....+\sqrt{a_{n-1}}-\sqrt{a_n}}{-d}=\frac{\sqrt{a_1}-\sqrt{a_n}}{-d}=\frac{\sqrt{a_n}-\sqrt{a_1}}{-d}\times\frac{\sqrt{a_n}+\sqrt{a_1}}{\sqrt{a_n}+\sqrt{a_1}}=\frac{a_n-a_1}{d(\sqrt{a_n}+\sqrt{a_1})}=\frac{(n-1)d}{d(\sqrt{a_n}+\sqrt{a_1})}=\frac{n-1}{\sqrt{a_n}+\sqrt{a_1}}$$
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