Answer:
a) 128 b) 512
Step-by-step explanation:
The key to solve this, is to think of it as a Geometric Sequence. Since, there is a constant ratio (q), each time the sheet of paper is cut, it is divided into two pieces or 1/2.
[tex]{\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16},\frac{1}{32},\frac{1}{64},\frac{1}{128},\frac{1}{256},\frac{1}{512}...[/tex]
If after the 1st cut there are two pieces i.e. 1/2
a) After the 7th cut, how many of the smallest pieces of paper are there?
You can either count the sequence above: 1/128 then 128 pieces or apply the recursive formula for any Term of a Geometric Sequence;
[tex]a_{n}=a_{1}* q^{n-1}\\ a_{7}=\frac{1}{2}*(\frac{1}{2})^{6} \\ a_{7}=\frac{1}{128}[/tex]
The Denominator indicates the smallest pieces of paper: 128
b) After the 9th cut, how many of the smallest pieces of paper are there?
Similarly to the item a: 1/512 then 512
[tex]a_{n}=a_{1}* q^{n-1}\\ a_{9}=\frac{1}{2}*(\frac{1}{2})^{8} \\ a_{9}=\frac{1}{512}[/tex]
The Denominator indicates the smallest pieces of paper after the Ninth cut: 512
