4

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


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!

Kenta S
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ronenp88
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  • Is this really true for any $a_i$. It seems to be false for instance if $a_i=\sqrt(i)$ (and $n=3)$. But it seems to be true (the answer below probably uses this) if $a_{i+1}=a_i+d$ for all $i$. – Jens Schwaiger Apr 26 '18 at 11:05

1 Answers1

7

Just rationalise the terms. The denominator becomes same for all, $d$, that is the common difference of AP:

$$S = \sum_{k = 1}^{n-1}\frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{d} = \frac{\sqrt{a_{n}}-\sqrt{a_1}}{d} = \frac{n-1}{\sqrt{a_{n}}+\sqrt{a_1}}$$

jonsno
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