0
$\begingroup$

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

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

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?

$\endgroup$
1

2 Answers 2

Reset to default
2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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

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

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

Terms in the middle cancels out

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

Which is what you observed.

$\endgroup$
2
  • $\begingroup$ I still don't get it, as far as the first equation you've written is concerned, why is it true? $\endgroup$
    – Pier
    May 22, 2020 at 13:06
  • $\begingroup$ @Pier $\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n(n+1)}-\frac{n}{n(n+1)}=\frac{1}{n(n+1)}$ can be proven by the reveresed direction. To "intuitively see how to find the RHS", look up "partial fraction decomposition" $\endgroup$
    – Gareth Ma
    May 22, 2020 at 13:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .