How can I show that if $(\chi_{n})$ is a aritmetic progression, then: $$\frac{1}{ \sqrt{\chi_{1}} + \sqrt{\chi_{2}} } + \frac{1}{ \sqrt{\chi_{2}} + \sqrt{\chi_{3}} } + \cdots + \frac{1}{ \sqrt{\chi_{n-1}} + \sqrt{\chi_{n}} } = \frac{n-1}{ \sqrt{\chi_{1}} + \sqrt{\chi_{n}} } $$
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Hints:
- $x_{k+1} = x_k + d$ with constant $d \Rightarrow x_n - x_1 = (n-1)d$
- $x_k - x_{k+1}=(\sqrt{x_k}-\sqrt{x_{k+1}})(\sqrt{x_k}+\sqrt{x_{k+1}})$
- Use this to transform the given sum into a telescoping one.
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$ \frac{1}{ \sqrt{\chi_{1}} + \sqrt{\chi_{2}} } + \frac{1}{ \sqrt{\chi_{2}} + \sqrt{\chi_{3}} } + \cdots + \frac{1}{ \sqrt{\chi_{n-1}} + \sqrt{\chi_{n}} } = \frac{\sqrt{\chi_{2}} - \sqrt{\chi_{1}} }{ \chi_2-\chi_1 } + \frac{\sqrt{\chi_{3}} - \sqrt{\chi_{2}} }{ \chi_3-\chi_2 } + \cdots + \frac{\sqrt{\chi_n}-\sqrt{\chi_{n-1}}}{ \chi_{n}-\chi_{n-1} } $
But $\chi_i-\chi_{i-1}=d$, the common difference of the arithmetic sequence.
$\frac{\sqrt{\chi_{2}} - \sqrt{\chi_{1}} }{ \chi_2-\chi_1 } + \frac{\sqrt{\chi_{3}} - \sqrt{\chi_{2}} }{ \chi_3-\chi_2 } + \cdots + \frac{\sqrt{\chi_n}-\sqrt{\chi_{n-1}}}{ \chi_{n}-\chi_{n-1} }=\frac{1}{d}\{\sqrt{\chi_{2}} - \sqrt{\chi_{1}}+\sqrt{\chi_{3}} - \sqrt{\chi_{2}}+\cdots+\sqrt{\chi_{n}} - \sqrt{\chi_{n-1}}\}$
$=\frac{\sqrt{\chi_{n}} - \sqrt{\chi_{1}}}{d}$
But since $\chi_n=\chi_1+(n-1)\cdot d$, we have $\frac{1}{d}=\frac{n-1}{\chi_n-\chi_1}$
Hence $\frac{\sqrt{\chi_{n}} - \sqrt{\chi_{1}}}{d}=\frac{(n-1)\cdot {\left(\sqrt{\chi_{n}} - \sqrt{\chi_{1}}\right)} }{\chi_n-\chi_1}=\frac{n-1}{\sqrt{\chi_{1}} + \sqrt{\chi_{n}}}$
