Series problem 3 by Dhaval Furia

Since $a_1\,,\,a_2\,,\,a_3\,,\,\cdots a_{n+1}$ are in arithmetic progression so their general term can be represented as:

$a_k = a_1 + (k-1)d$ where $d$ is the common difference between two consecutive terms of progression.

We have

$a_2 - a_1 = d\,,\,a_3 - a_2 = d\,,\,a_4 - a_3 = d$ and in general

$a_{k+1} - a_k = d \forall k = 1$ to $n$

Now coming to our expression,

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

There are total $n$ terms. Multiplying $\sqrt{a_{k+1}} - \sqrt{a_k}$ in both numerator and denominator of $k$ th term $(k = 1$ to $n)$ of the above expression we get

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

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

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

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

$\displaystyle = \frac{\sqrt{a_{n+1}} - \sqrt{a_1}}{d}$

Multiplying $\sqrt{a_{n+1}} + \sqrt{a_1}$ in both numerator and denominator we get

$\displaystyle \frac{(\sqrt{a_{n+1}} - \sqrt{a_1})(\sqrt{a_{n+1}} + \sqrt{a_1})}{d(\sqrt{a_{n+1}} + \sqrt{a_1})}$

$\displaystyle = \frac{a_{n+1} - a_1}{d(\sqrt{a_{n+1}} + \sqrt{a_1})}$

$\displaystyle = \frac{a_1 + nd - a_1}{d(\sqrt{a_{n+1}} + \sqrt{a_1})}$

$\displaystyle = \frac{nd}{d(\sqrt{a_{n+1}} + \sqrt{a_1})}$

$\displaystyle = \frac{n}{\sqrt{a_{n+1}} + \sqrt{a_1}}$