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
 A: Assume that $a_i = a_1+(i-1)\delta$ for all $i > 0$ and some $\delta$.  Proceed by induction.  Let
$$
S_k = \frac{1}{\sqrt{a_1}+\sqrt{a_2}}+
\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\cdots+
\frac{1}{\sqrt{a_{k-1}}+\sqrt{a_k}}
$$
Observe that for $k = 2$, we have
$$
S_2 = \frac{1}{\sqrt{a_1}+\sqrt{a_2}}
$$
trivially.  Next, suppose that
$$
S_k = \frac{k-1}{\sqrt{a_1}+\sqrt{a_k}}
$$
Then, keeping in mind that $a_k = a_1+(k-1)\delta$,
\begin{align}
S_{k+1}
    & = \frac{1}{\sqrt{a_1}+\sqrt{a_2}}+
        \frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\cdots+
        \frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}} \\
    & = \frac{k-1}{\sqrt{a_1}+\sqrt{a_k}} + \frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}} \\
    & = \frac{k-1}{\sqrt{a_1}+\sqrt{a_1+(k-1)\delta}}
      + \frac{1}{\sqrt{a_k}+\sqrt{a_k+\delta}} \\
    & = \frac{(k-1)(\sqrt{a_1+(k-1)\delta}-\sqrt{a_1})}{(k-1)\delta}
      + \frac{\sqrt{a_k+\delta}-\sqrt{a_k}}{\delta} \\
    & = \frac{\sqrt{a_k}-\sqrt{a_1}}{\delta}
      + \frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{\delta} \\
    & = \frac{\sqrt{a_{k+1}}-\sqrt{a_1}}{\delta} \\
    & = \frac{\sqrt{a_1+k\delta}-\sqrt{a_1}}{\delta} \\
    & = \frac{k\delta}{\delta(\sqrt{a_1+k\delta}+\sqrt{a_1})} \\
    & = \frac{k}{\sqrt{a_1}+\sqrt{a_{k+1}}}
\end{align}
For instance,
\begin{align}
\frac{1}{\sqrt{1}+\sqrt{25}}+\frac{1}{\sqrt{25}+\sqrt{49}}
    & = \frac{1}{1+5}+\frac{1}{5+7} \\
    & = \frac{1}{6}+\frac{1}{12} \\
    & = \frac{3}{12} \\
    & = \frac{2}{8} \\
    & = \frac{3-1}{\sqrt{1}+\sqrt{49}}
\end{align}
A: $$\frac{\sqrt{a_1}-\sqrt{a_2}}{{a_1} + {a_2}}+\frac{\sqrt{a_2}-\sqrt{a_3}}{{a_2}-{a_3}}+....+\frac{\sqrt{a_{n-1}}-\sqrt{a_n}}{a_{n-1}-{a_n}}=\frac{\sqrt{a_1}-\sqrt{a_2}}{-d}+\frac{\sqrt{a_2}-\sqrt{a_3}}{-d}+....+\frac{\sqrt{a_{n-1}}-\sqrt{a_n}}{-d}=\frac{\sqrt{a_1}-\sqrt{a_2}+\sqrt{a_2}-\sqrt{a_3}+....+\sqrt{a_{n-1}}-\sqrt{a_n}}{-d}=\frac{\sqrt{a_1}-\sqrt{a_n}}{-d}=\frac{\sqrt{a_n}-\sqrt{a_1}}{-d}\times\frac{\sqrt{a_n}+\sqrt{a_1}}{\sqrt{a_n}+\sqrt{a_1}}=\frac{a_n-a_1}{d(\sqrt{a_n}+\sqrt{a_1})}=\frac{(n-1)d}{d(\sqrt{a_n}+\sqrt{a_1})}=\frac{n-1}{\sqrt{a_n}+\sqrt{a_1}}$$
