# Could you please prove the following equation from Arithmetic Progressions?

## If ${a}_{1}$, ... , ${a}_{n}$ are in AP where, ${a}_{i} > 0$ for all $i$, then show that $\frac{1}{\sqrt{{a}_{1}} + \sqrt{{a}_{2}}} + \frac{1}{\sqrt{{a}_{2}} + \sqrt{{a}_{3}}} + \ldots . + \frac{1}{\sqrt{{a}_{n} - 1} + \sqrt{{a}_{n}}} = \frac{n - 1}{\sqrt{{a}_{1}} + \sqrt{{a}_{n}}}$.

Feb 25, 2018

See below.

#### Explanation:

If the sequence ${a}_{k}$ is an AP then

${a}_{k + 1} = {a}_{k} + \Delta$ so

$\frac{1}{\sqrt{{a}_{k}} + \sqrt{{a}_{k + 1}}} = \frac{\sqrt{{a}_{k + 1}} - \sqrt{{a}_{k}}}{{a}_{k + 1} - {a}_{k}} = \frac{\sqrt{{a}_{k + 1}} - \sqrt{{a}_{k}}}{\Delta}$

and then

${\sum}_{k = 1}^{n - 1} \frac{1}{\sqrt{{a}_{k}} + \sqrt{{a}_{k + 1}}} = \frac{1}{\Delta} \left({\sum}_{k = 1}^{n - 1} \sqrt{{a}_{k + 1}} - {\sum}_{k = 1}^{n - 1} \sqrt{{a}_{k}}\right) = \frac{1}{\Delta} \left(\sqrt{{a}_{n}} - \sqrt{{a}_{1}}\right)$

but

$\frac{1}{\Delta} \left(\sqrt{{a}_{n}} - \sqrt{{a}_{1}}\right) = \frac{1}{\Delta} \left(\sqrt{{a}_{n}} - \sqrt{{a}_{1}}\right) \left(\frac{\sqrt{{a}_{n}} + \sqrt{{a}_{1}}}{\sqrt{{a}_{n}} + \sqrt{{a}_{1}}}\right) = \frac{1}{\Delta} \frac{{a}_{n} - {a}_{1}}{\sqrt{{a}_{n}} + \sqrt{{a}_{1}}} = \frac{1}{\Delta} \frac{\left(n - 1\right) \Delta}{\sqrt{{a}_{n}} + \sqrt{{a}_{1}}} = \frac{n - 1}{\sqrt{{a}_{n}} + \sqrt{{a}_{1}}}$