Question

An insect is on the 0 point of a number line, hopping towards 1. She covers half the distance from her current location to 1 with each hop. So, she will be at ${\displaystyle \frac{1}{2}}$ after one hop, ${\displaystyle \frac{3}{4}}$ after two hops, and so on.

1. Make a table showing the insect’s location for the first 10 hops.
2. Where will the insect be after n hops?
3. Will the insect ever get to 1? Explain.

Answer 1

a. On the basis of given information in the question, we can arrange the following table which shows the insect’s location for the first 10 hops.

Number of hops	Distance covered	Distance left	Distance covered
1.	${\displaystyle \frac{1}{2}}$	${\displaystyle \frac{1}{2}}$	${\displaystyle 1-\frac{1}{2}}$
2.	${\displaystyle \frac{1}{2}\left(\frac{1}{2}\right)+\frac{1}{2}}$	${\displaystyle \frac{1}{4}}$	${\displaystyle 1-\frac{1}{4}}$
3.	${\displaystyle \frac{1}{2}\left(\frac{1}{4}\right)+\frac{3}{4}}$	${\displaystyle \frac{1}{8}}$	${\displaystyle 1-\frac{1}{8}}$
4.	${\displaystyle \frac{1}{2}\left(\frac{1}{8}\right)+\frac{7}{8}}$	${\displaystyle \frac{1}{16}}$	${\displaystyle 1-\frac{1}{16}}$
5.	${\displaystyle \frac{1}{2}\left(\frac{1}{16}\right)+\frac{15}{16}}$	${\displaystyle \frac{1}{32}}$	${\displaystyle 1-\frac{1}{32}}$
6.	${\displaystyle \frac{1}{2}\left(\frac{1}{32}\right)+\frac{31}{32}}$	${\displaystyle \frac{1}{64}}$	${\displaystyle 1-\frac{1}{64}}$
7.	${\displaystyle \frac{1}{2}\left(\frac{1}{64}\right)+\frac{63}{64}}$	${\displaystyle \frac{1}{128}}$	${\displaystyle 1-\frac{1}{128}}$
8.	${\displaystyle \frac{1}{2}\left(\frac{1}{128}\right)+\frac{127}{128}}$	${\displaystyle \frac{1}{256}}$	${\displaystyle 1-\frac{1}{256}}$
9.	${\displaystyle \frac{1}{2}\left(\frac{1}{256}\right)+\frac{255}{256}}$	${\displaystyle \frac{1}{512}}$	${\displaystyle 1-\frac{1}{512}}$
10.	${\displaystyle \frac{1}{2}\left(\frac{1}{512}\right)+\frac{511}{512}}$	${\displaystyle \frac{1}{1024}}$	${\displaystyle 1-\frac{1}{1024}}$

b. If we see the distance covered in each hops

Distance covered in 1^st hop = ${\displaystyle 1-\frac{1}{2}}$

Distance covered in 2^nd hops = ${\displaystyle 1-\frac{1}{4}}$

Distance covered in 3^rd hops = ${\displaystyle 1-\frac{1}{8}}$

Distance covered in n hops = ${\displaystyle 1-{\left(\frac{1}{2}\right)}^n}$

c. No, because for reaching ${\displaystyle 1,{\left(\frac{1}{2}\right)}^n}$ has to be zero for some finite n which is not possible.