How to simply this fraction with irrational denominators?

Question

How to simplify?

$\frac{1}{1+\sqrt{3}} + \frac{1}{\sqrt{3}+\sqrt{5}} + \frac{1}{\sqrt{5}+\sqrt{7}} \frac{1}{\sqrt{7}+3}$

Answer 1

Hint. If you rationalise the denominator of each term, you always obtain the same denominator and, in the numerator, a telescoping sum.

Answer 2

If it is $\frac{1}{1+\sqrt{3}} + \frac{1}{\sqrt{3}+\sqrt{5}} + \frac{1}{\sqrt{5}+\sqrt{7}}\color{red}+ \frac{1}{\sqrt{7}+3}$ then

$$=\frac{1}{1+\sqrt{3}}\frac{1-\sqrt3}{1-\sqrt3} + \frac{1}{\sqrt{3}+\sqrt{5}}\frac{\sqrt3-\sqrt5}{\sqrt3-\sqrt5} + \frac{1}{\sqrt{5}+\sqrt{7}}\frac{\sqrt5-\sqrt7}{\sqrt5-\sqrt7}+ \frac{1}{\sqrt{7}+3}\frac{\sqrt7-3}{\sqrt7-3}$$

$$=\frac{1-\sqrt3}{1-3}+\frac{\sqrt{3}-\sqrt{5}}{3-5}+\frac{\sqrt5-\sqrt7}{5-7}+\frac{\sqrt7-3}{7-9}$$

$$= \frac{1-\sqrt3}{-2}+\frac{\sqrt{3}-\sqrt{5}}{-2}+\frac{\sqrt5-\sqrt7}{-2}+\frac{\sqrt7-3}{-2}$$

$$=-\frac{1}{2}(1-\sqrt3+\sqrt3-\sqrt5+\sqrt5-\sqrt7+\sqrt7-3)$$

$$=-\frac{1}{2}(1-3)$$

$$=-\frac{1}{2}\times -2$$

$$=1$$

Answer 3

If b>a $\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{\sqrt{b}-\sqrt{a}}{b-a}$

Therefore in your question their all down parts equal two with that formula

Then $\frac{\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+\sqrt{9}-\sqrt{7}...}{2}$

As you see there is pattern... Pattern is that for 4 number --> (num_curr)(-num_pre)(num_next)(-num_curr).

(biggest(num_next) - smallest(num_pre))/2 (for your example (3-1)/2=2/2=1)