<s>[INST] Summation of series with $r^{th}$ term $\frac{1}{\sqrt{a+rx} +\sqrt{a+(r-1)x}} $ 
Find the summation of series with $r^{th}$ term $t_r$ given by $\displaystyle \frac{1}{\sqrt{a+rx} +\sqrt{a+(r-1)x}} $     

Working:
$$t_r=\frac{1}{\sqrt{a+rx} +\sqrt{a+(r-1)x}}$$  
$$t_r=\frac{\sqrt{a+rx} -\sqrt{a+(r-1)x}}{a+rx-(a+(r-1)x)}$$      
$$t_r=\frac{\sqrt{a+rx} -\sqrt{a+(r-1)x}}{x}$$     
How do I calculate the summation $S_n$, from the $r^{th}$ term?
 [/INST] Continue from your working $$S_{n} = \sum^{n}_{r=1}t_{r} = \frac{1}{x}\sum^{n}_{r=1}\bigg[\sqrt{a+rx}-\sqrt{a+(r-1)x}\bigg]$$
Now Using Telescopis Sum (expanding summation.)
$$S_{n} = \frac{1}{x}\bigg[\left(\sqrt{a+x}-\sqrt{a}\right)+(\sqrt{a+2x}-\sqrt{a+x})+\cdots \cdots +(\sqrt{a+nx}-\sqrt{a+(n-1)x)}\bigg]$$
So we get $$S_{n} = \frac{1}{x}(\sqrt{a+nx}-\sqrt{a}) = \frac{(\sqrt{a+nx}-\sqrt{a})\cdot (\sqrt{a+nx}+\sqrt{a})}{\sqrt{a+nx}+\sqrt{a}} = \frac{n}{\sqrt{a}+\sqrt{a+nx}}$$
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