Chapter 7, Problem 32T

Solution

i.

To state: The representation by n for each quilt and also what represents.

  

The number of blue squares is represented by $a_n$ and n represents number of rows and columns.

Given information:

The given pattern of checkerboards quilt is:

  

Explanation:

Consider the given image of the quilt.

  

Now observe the pattern. The first quilt is not divided and the second quilt has 2 rows and 2 columns.

The third quilt has 3 rows and 3 columns and the fourth quilt has 4 rows and 4 columns and so on.

It can be seen that n represents the number of rows and columns in which the quilt is divided.

Now observe the pattern again, the first quilt is blue $a_1=1$, the second quilt has 2 blue squares and the third quilt has 5 blue squares $a_3=5$ and so on.

Therefore, it can be said that represents the number of blue squares.

ii.

To state: A table that shows n and for $n=1,2,3,4,5,6,7,\text{and }8$ .

n	1	2	3	4	5	6	7	8
	1	2	5	8	13	18	25	32

Given information:

The given values are $n=1,2,3,4,5,6,7,\text{and }8$ .

Explanation:

From the pattern it can be said that:

For $n=1$ the term is $a_1=1$ .

For $n=2$ the term is $a_2=2$ .

For $n=3$ the term is $a_3=5$ .

If the number n is even then the number of blue squares: $a_n=\frac{n^2}{2}$

If the number n is odd then the number f blue square: $a_n=\frac{n^2+1}{2}$

Now make a table showing n and

n	1	2	3	4	5	6	7	8
	1	2	5	$\begin{array}{c}{a}_4=\frac{4^2}{2}\\ =8\end{array}$	$\begin{array}{c}{a}_5=\frac{5^2+1}{2}\\ =13\end{array}$	$\begin{array}{c}{a}_6=\frac{6^2}{2}\\ =18\end{array}$		$\begin{array}{c}{a}_8=\frac{8^2}{2}\\ =32\end{array}$

iii.

To state: The values of for $n=1,2,3,4,5,6,7,\text{and }8$ using the rule $a_n=\frac{n^2}{2}+\frac{1}{4}\left[1-{\left(-1\right)}^n\right]$ .

The rule $a_n=\frac{n^2}{2}+\frac{1}{4}\left[1-{\left(-1\right)}^n\right]$ is for calculating the number of blue squares.

Given information:

The given rule is

Explanation:

Consider the rule: $a_n=\frac{n^2}{2}+\frac{1}{4}\left[1-{\left(-1\right)}^n\right]$

Now substitute the values of n as $1,2,3,4,5,6,7,\text{and }8$:

  $\begin{array}{l}{a}_1=\frac{1^2}{2}+\frac{1}{4}\left[1-{\left(-1\right)}^1\right],a_1=\frac{1}{2}+\frac{1}{4}\cdot 2,a_1=1\hfill \\ a_2=\frac{2^2}{2}+\frac{1}{4}\left[1-{\left(-1\right)}^2\right],a_2=2+\frac{1}{4}\cdot 0,a_2=2\hfill \end{array}$

  $\begin{array}{l}{a}_3=\frac{3^2}{2}+\frac{1}{4}\left[1-{\left(-1\right)}^3\right],a_3=\frac{9}{2}+\frac{1}{4}\cdot 2,a_3=5\hfill \\ a_4=\frac{4^2}{2}+\frac{1}{4}\left[1-{\left(-1\right)}^4\right],a_4=8+\frac{1}{4}\cdot 0,a_4=8\hfill \end{array}$

  $\begin{array}{l}{a}_5=\frac{5^2}{2}+\frac{1}{4}\left[1-{\left(-1\right)}^5\right],a_5=\frac{25}{2}+\frac{1}{4}\cdot 2,a_5=13\hfill \\ a_6=\frac{6^2}{2}+\frac{1}{4}\left[1-{\left(-1\right)}^6\right],a_6=18+\frac{1}{4}\cdot 0,a_6=18\hfill \end{array}$

  $\begin{array}{l}{a}_7=\frac{7^2}{2}+\frac{1}{4}\left[1-{\left(-1\right)}^7\right],a_5=\frac{49}{2}+\frac{1}{4}\cdot 2,a_5=25\hfill \\ a_8=\frac{8^2}{2}+\frac{1}{4}\left[1-{\left(-1\right)}^8\right],a_8=32+\frac{1}{4}\cdot 0,a_8=32\hfill \end{array}$

It can be seen that the values of from rule $a_n=\frac{n^2}{2}+\frac{1}{4}\left[1-{\left(-1\right)}^n\right]$ and that of from part (b) are same.

It can be concluded that the rule is for calculating the number of blue squares.