How to answer this kind of questions

Question

How can I answer this kind of questions?

I should prove the following

$$\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \frac{1}{\sqrt{a_3} + \sqrt{a_4}} + ... + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}} = \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}$$

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Sorry for my poor English

Sorry I edited the picture and it removed , I will replace it — , Oct 05 '15 at 19:45
No problem. For the future, there here is an explanation of how to type math on this website. — Ben Grossmann, Oct 05 '15 at 19:47
Is there a rule for the $a_n$? This isn't true in general for any sequence $a_1, a_2, \ldots, a_n$. For instance $\frac{1}{\sqrt{1}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{9}} \not= \frac{2}{\sqrt{1}+\sqrt{9}}$. — Brian Tung, Oct 05 '15 at 19:54
I'm assuming the $a_i$ form an arithmetic sequence, based on the proposition? — Brian Tung, Oct 05 '15 at 19:57

score 1 · Answer 1 · answered Oct 05 '15 at 20:18

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}

score 0 · Accepted Answer · answered Oct 09 '15 at 08:42

$$\frac{\sqrt{a_1}-\sqrt{a_2}}{{a_1} + {a_2}}+\frac{\sqrt{a_2}-\sqrt{a_3}}{{a_2}-{a_3}}+....+\frac{\sqrt{a_{n-1}}-\sqrt{a_n}}{a_{n-1}-{a_n}}=\frac{\sqrt{a_1}-\sqrt{a_2}}{-d}+\frac{\sqrt{a_2}-\sqrt{a_3}}{-d}+....+\frac{\sqrt{a_{n-1}}-\sqrt{a_n}}{-d}=\frac{\sqrt{a_1}-\sqrt{a_2}+\sqrt{a_2}-\sqrt{a_3}+....+\sqrt{a_{n-1}}-\sqrt{a_n}}{-d}=\frac{\sqrt{a_1}-\sqrt{a_n}}{-d}=\frac{\sqrt{a_n}-\sqrt{a_1}}{-d}\times\frac{\sqrt{a_n}+\sqrt{a_1}}{\sqrt{a_n}+\sqrt{a_1}}=\frac{a_n-a_1}{d(\sqrt{a_n}+\sqrt{a_1})}=\frac{(n-1)d}{d(\sqrt{a_n}+\sqrt{a_1})}=\frac{n-1}{\sqrt{a_n}+\sqrt{a_1}}$$

How to answer this kind of questions

2 Answers2