Prove that $\sum_{k=2}^{n}\{\sqrt{a_{k-1}}+\sqrt{a_{k}}\}^{-1}=\frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}}$

Question

If $a_{1}, a_{2}, a_{3}, \dots, a_{n}$ are consecutive terms of an arithmetic progression (None of which are $0$) , prove that $\sum_{k=2}^{n}\{\sqrt{a_{k-1}}+\sqrt{a_{k}}\}^{-1}=\frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}}$

I started out doing this problem by approaching it using mathematical induction. If my calculations are right, it must be proved through induction that

$\frac{k-1}{\sqrt a_{1}+\sqrt{a_{n}}}+\frac{1}{\sqrt{a_{k}}+\sqrt{a_{k-1}}}=\frac{k}{\sqrt{a_{1}}+\sqrt{a_{k-1}}}$

Trying to simplify this gives me a very long expression. I assume that we have to use the fact that the terms are in AP. Any help is appreciated.

Answer 1

Suppose that $d$ is the common difference. Rationalizing:

$$\frac{1}{\sqrt{a_{k-1}} + \sqrt{a_k}} = - \frac{\sqrt{a_{k-1}} - \sqrt{a_k}}{d}$$

Therefore:

$$\begin{aligned} \sum_{k=2}^{n}\frac{1}{\sqrt{a_{k-1}}+\sqrt{a_{k}}} &= \sum_{k=2}^n \left(- \frac{\sqrt{a_{k-1}} - \sqrt{a_k}}{d}\right)\\ &=-\frac{\sqrt{a_1} - \require{cancel}\cancel{\sqrt{a_2}}}{d}-\frac{\require{cancel}\cancel{\sqrt{a_2}} - \require{cancel}\cancel{\sqrt{a_3}}}{d}-\ldots-\frac{\require{cancel}\cancel{\sqrt{a_{n-1}}} - \sqrt{a_n}}{d} \\ &=-\frac{\sqrt{a_1} - \sqrt{a_n}}{d} \\ &= \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}. \end{aligned}$$

The last equality is equivalent with $a_n=a_1+(n-1)d$.