I have found the sketch of this problem that is shown in the figure below. So let's solve this problem. We know the following expression:
[tex]t=4(b-1) + 10 \\ \\ t:Colored \ tiles \\ b:Border \ number \\ with \ b \geq 1[/tex]
a. Explain why Fred's expression is correct.
The expression is correct and we can prove it by analyzing each border, that is, border 1, border 2, border 3 and so on.
For Border 1, [tex]b=1[/tex]
[tex]t=4(1-1)+10=10[/tex]
For Border 1 there are 10 tiles colored in red
For Border 2, [tex]b=2[/tex]
[tex]t=4(2-1)+10=14[/tex]
For Border 2 there are 14 tiles colored in red
For Border 3, [tex]b=3[/tex]
[tex]t=4(3-1)+10=18[/tex]
For Border 3 there are 18 tiles colored in red
So in conclusion, Fred has written a correct formula for this problem. Note that each border has a constant value of 10 tiles colored in red. As the number of border is increasing the number of tiles is also increasing in steps of four, that is:
[tex]For \ border \ 1: \\ t=0+10 \\ \\ For \ border \ 2: \\ t=4+10 \\ \\ For \ border \ 3: \\ t=4+4+10 \\ \\ For \ border \ 3: \\ t=4+4+4+10 \\ \\ For \ border \ b: \\ t=4(b-1)+10 [/tex]
b. Is Emma’s statement correct? Explain your reasoning.
Emma is reasoning incorrectly so her statement is not correct. She has incorrectly assumed that the 4 in Fred’s formula came from the number of tiles in the
beginning row but this really comes from the number of corners in the diagram
itself regardless of the number of tiles in the beginning row. Therefore, there will always be four corners. The formula for her must be:
[tex]t=4(b-1)+12[/tex]
There will be a constant value of 12 tiles colored in red for her because starting with border 1 there will be 5 tiles in red at the top of the five empty tiles and another 5 tiles in red at the bottom plus one red tile in each end.
c. If Emma starts with a row of n tiles, what should the expression be?
The
number of tiles needed for each corner will remain the same regardless of the
number of tiles in the original row, that is the term:
[tex]4(b-1)[/tex]
always remains. So we need to find another term, that is:
[tex]If \ n=4 \\ t=4(b-1)+10 \\ \\ If \ n=4 \\ t=4(b-1)+12[/tex]
So we can rewrite that in this way:
[tex]If \ n=4 \\ t=4(b-1)+2(4+1) \\ \\ If \ n=4 \\ t=4(b-1)+2(5+1)[/tex]
In general form:
[tex]For \ n: \\ \\ \boxed{t=4(b-1)+2(n+1)}[/tex]
