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

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?

• Did you try the old rationalizing the denominator trick? I don't know if it will work, but it's the obvious next thing to try. Commented Mar 20, 2021 at 15:39
• @LeeMosher didn't think of that, thanks a lot, it works with that Commented Mar 20, 2021 at 15:40
• Commented Mar 20, 2021 at 17:05

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