Problems

Above are three example solutions for finding the number of squares in the nth pattern. 

A) We first notice that the top and bottom rows of the pattern are each 2n squares long, leaving a rectangle between them. The rectangle is n squares wide, but the height changes by two each time. We see in the first four patterns that the height is consecutive odd numbers, starting at 1, growing to 3, then 5, and 7. Odd numbers are an arithmetic sequence starting at 1 with common difference 2. This expression is not simplified in these example solutions but will be simplified to 2n - 1 in subsequent solutions.

B) Maybe we notice that the top and bottom rows are 4 sets of n squares total. This results in a slightly different expression for the first term because we want to represent counting four groups of n rather than s groups of 2n.

C) In the third example, we chop off the 2 "arms" that are n squares long on the right. This increases the height of the rectangle by 2 so our odd numbered heights start at 3 instead of 1. Instead of writing this as the arithmetic sequence 3 + 2(n - 1) we simplify it to 1 + 2n.

Algebraic Equivalences:

Above are three example solutions for finding the number of squares in the nth pattern. 

A) If we notice the top and bottom rows each have n squares in them, we are left with a square between them. The side lengths of the square are 1 less than n so we can add the first 2 rows to the area of the square.

B) Cutting off each of the squares on the bottom left and top right of the figure, we are left with a rectangle. The rectangle has a width that is 1 less than n and a height that is 1 more than n. We can count all the squares by adding the area of the rectangle to the 2 poking out on the sides.

C) Instead of adding all the areas together, here we subtract the "negative space". If we imagine a square containing our figure, it is a square with side lengths that are 1 longer than n. The two strips of negative space are each made up of n squares. Here we obtain our expression by subtracting the 2n squares from the area of the n + 1 square. Can you find a solution for Pattern 1 by subtracting negative space?

Algebraic Equivalences:

Above are four example solutions for finding the number of squares in the nth pattern. 

A) We can view this pattern as having 4 arms surrounding a center square. The arms are all 1 square shorter than n so we can add 1 to 4(n - 1).

B) If we prefer to see our 4 arms each having n squares, then we have to take into account that the square at the center is being counted 4 times instead of 1. To adjust for our overcounting, we subtract 3 from the sum of the 4 arms.

C) Sometimes instead of seeing 4 groups of n - 1 squares, it is easier to see concentric "circles" of 4 squares. There are n - 1 circles of 4 squares, until the center square where there is just 1. Algebraically, the only difference between this and solution A) is the order of the factors in the first term. However, the picture looks different because we are viewing the groupings and using a different strategy to count the squares.

D) Do you see the pyramids forming in the negative space in each figure? If we shift each pyramid on the left and right down one unit and the top down two, we form a rectangle that is 2n - 1 squares wide and 2n - 3 with the tops of the bottom and top pyramid overlapping at one square. We can subtract the number of squares making up this rectangle from the square containing the whole pattern, along with the 1 square of overlap.

Algebraic Equivalences:

Above are two example solutions for finding the number of squares in the nth pattern. 

A) If you've successfully described Pattern 3 then you might notice that the same pattern is embedded in a larger square. We learned several ways in our practice for finding the boundary of a square; even though the boundary of this one is growing by 2 between each figure starting at 3, we can still modify one of our strategies to work here. Once we have expressions for each of the patterns, we can add them together to find the total number of squares in each figure.

B) In this strategy, we extend the inner pattern all the way to the outer edge of the figure and count up the number of squares between each corner on the boundary.

Algebraic Equivalences: