Find the sum of $\frac{1}{\sqrt1+\sqrt3} + \frac{1}{\sqrt3+\sqrt5} + \frac{1}{\sqrt5+\sqrt7} + ... \frac{1}{\sqrt79+\sqrt81}$ Find the sum of $\frac{1}{\sqrt1+\sqrt3} + \frac{1}{\sqrt3+\sqrt5} + \frac{1}{\sqrt5+\sqrt7} + ... \frac{1}{\sqrt{79}+\sqrt{81}}$
I've thought about multiplying every fraction by 1, but like this $\frac{1} {\sqrt1+\sqrt3} *\frac{\sqrt1+\sqrt3}{\sqrt1+\sqrt3}$, $\frac{1} {\sqrt3+\sqrt5} *\frac{\sqrt3+\sqrt5}{\sqrt3+\sqrt5}...$
After multiplying I end up with $\frac{\sqrt1+\sqrt3} {(\sqrt1+\sqrt3)^2} + \frac{\sqrt3+\sqrt5} {(\sqrt3+\sqrt5)^2}  ...$ but I don't have any more ideas than this, am I on the right path?
I apologize for the question's simplicity, but I'm a younger math enthusiast so questions like this are harder for me.
 A: You have the right idea, except you should rationalize the denominators by multiplying the numerators and denominators by the fractions denominators' conjugates, i.e., with the "+" replaced by "-". For example,
$$\begin{equation}\begin{aligned}
\frac{1}{\sqrt{1} + \sqrt{3}} & = \left(\frac{1}{\sqrt{1} + \sqrt{3}}\right)\left(\frac{\sqrt{1} - \sqrt{3}}{\sqrt{1} - \sqrt{3}}\right) \\
& = \frac{\sqrt{1} - \sqrt{3}}{1 - 3} \\
& = \frac{\sqrt{1} - \sqrt{3}}{-2}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
You will find all of the denominators become $-2$, and the sum of the numerators form a telescoping series, so almost all terms cancel. I'll let you finish the rest.
A: Hint: $\frac 1{\sqrt{n} + \sqrt{n+2}} \cdot \frac {\sqrt{n+2} - \sqrt{n}}{\sqrt{n+2} - \sqrt{n}} = \frac{\sqrt{n+2} - \sqrt{n}}{2}$
A: $\frac{1}{\sqrt1+\sqrt3} + \frac{1}{\sqrt3+\sqrt5} + \frac{1}{\sqrt5+\sqrt7} + ... \frac{1}{\sqrt{79}+\sqrt{81}}=\frac{1}{\sqrt3+\sqrt1} *\frac{\sqrt3-\sqrt1}{\sqrt3-\sqrt1}+ \frac{1}{\sqrt5+\sqrt3}*\frac{\sqrt5-\sqrt3}{\sqrt5-\sqrt3} + \frac{1}{\sqrt7+\sqrt5}*\frac{\sqrt7-\sqrt5}{\sqrt7-\sqrt5} + ... \frac{1}{\sqrt{81}+\sqrt{79}} *\frac{\sqrt81-\sqrt79}{\sqrt81-\sqrt79}=\frac{\sqrt3-\sqrt1}{2}+\frac{\sqrt5-\sqrt3}{2}+\frac{\sqrt7-\sqrt5}{2}+...+\frac{\sqrt81-\sqrt79}{2} =(\frac{1}{2})((\sqrt3-\sqrt1)+(\sqrt5-\sqrt3)+(\sqrt7-\sqrt5)+...(\sqrt81-\sqrt79))=(\frac{1}{2})(\sqrt81-\sqrt1)=4$
So the problem is done.
