<s>[INST] Find $f(1)+\cdots+f(60)$ 
Let $f(x) = \dfrac{4x+\sqrt{4x^2-1}}{\sqrt{2x+1}+\sqrt{2x-1}}$. Find $f(1)+\cdots+f(60)$.

I considered rationalizing the denominator, but that seems to make the fraction more complicated. We get $$\dfrac{4x+\sqrt{4x^2-1}}{\sqrt{2x+1}+\sqrt{2x-1}} = \dfrac{1}{2}\left(4x+\sqrt{4x^2-1}\right)\left(\sqrt{2x+1}-\sqrt{2x-1}\right).$$ is there an easier way?
 [/INST] Using the fact that $(\sqrt{2x+1}+\sqrt{2x-1})^2=4x+2\sqrt{4x^2-1}$, you can produce :
\begin{align}
  \frac{4x+\sqrt{4x^2-1}}{\sqrt{2x+1}+\sqrt{2x-1}} &=
  \frac{(\sqrt{2x+1}+\sqrt{2x-1})^2-\sqrt{4x^2-1}}{\sqrt{2x+1}+\sqrt{2x-1}} \\
&= \sqrt{2x+1}+\sqrt{2x-1} - \frac{\sqrt{4x^2-1}}{\sqrt{2x+1}+\sqrt{2x-1}} \\ &= \sqrt{2x+1}+\sqrt{2x-1} - \frac{\sqrt{4x^2-1}(\sqrt{2x+1}-\sqrt{2x-1})}{2} \\ 
&= \sqrt{2x+1}+\sqrt{2x-1} - \frac{(2x+1)}{2}\sqrt{2x-1} + \frac{(2x-1)}{2}\sqrt{2x+1} \\ 
&= \frac{(2x+1)^{3/2}}{2} - \frac{(2x-1)^{3/2}}{2}
\end{align}
So your sum is telescopic, and you find :
$$\sum_{k=1}^{60} f(k) = \frac{(2\times 60+1)^{3/2}}{2} - \frac{(2\times 1-1)^{3/2}}{2} = 665$$
Note : I'm sure there's a much cleaner way to obtain this, but I don't see how :-)
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