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Find the value of the expression: $$\frac{1}{\sqrt{4}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{10}}+\cdots+\frac{1}{\sqrt{13}+\sqrt{16}}$$

After putting it into a calculator I worked out that it is equal to $\frac{2}{3}$. I tried to work it out on my own by saying $$\sum\limits_{k=1}^{5}\frac{1}{\sqrt{3k+1}+\sqrt{3k+4}},$$ must resemble somehow the telescopic identity: $$\frac{1}{m(m+1)}=\frac{1}{m}-\frac{1}{m+1},$$ but I couldn't find a way for it to become similar to it. I then said $$\sum\limits_{k=1}^{5}\frac{1}{\sqrt{3k+1}+\sqrt{3k+4}}=\sum\limits_{k=2}^{6}\frac{1}{\sqrt{3k-2}+\sqrt{3k+1}},$$ but that didn't bare crop either. Could you please explain to me how to solve this question?

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Hint You had the nice idea with the telescopic sum :

$$\frac{1}{\sqrt{3k+1}+\sqrt{3k+4}}=\frac{\sqrt{3k+1}-\sqrt{3k+4}}{(\sqrt{3k+1}-\sqrt{3k+4})(\sqrt{3k+1}+\sqrt{3k+4})}=\frac{\sqrt{3k+1}-\sqrt{3k+4}}{3k+1-3k-4}=\frac{\sqrt{3k+1}-\sqrt{3k+4}}{-3}$$

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