Explaining the pattern

Question

$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3} =\frac12 + \frac16 = \frac36+\frac16 = \frac46 = \boxed{\tfrac23}$

$\left(\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}\right) +\frac{1}{3\cdot 4} =\frac23 +\frac{1}{3\cdot 4} = \frac{2\cdot4}{3\cdot 4} + \frac{1}{3\cdot 4} =\frac{9}{12} = \boxed{\dfrac34}$

Every time I add a fraction whose numerator is 1, and whose denominator is the product of the first term of the denominator of the previous fraction + 1 x the second term of the denominator of the previous fraction If I keep going I'll have $\frac{4}{5}$, then $\frac{4}{5}$ and so on. Why is there this pattern?

Answer 1

Because $\frac{1}{n(n+1)} = \frac1n - \frac{1}{n+1}$.

Answer 2

$$\frac{1}{n\cdot (n+1)}=\frac{1}{n}-\frac{1}{n+1}$$

$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots + \frac{1}{n\cdot (n+1)}$$

$$= \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdots +\frac{1}{n}-\frac{1}{n+1}$$

Terms in the middle cancels out

$$=1-\frac{1}{n+1}=\frac{n}{n+1}$$

Which is what you observed.