Module 3 - Investigation Structure
TL;DR
To justify means to argue that the general rule is always true.
You can do this by:
using the geometry of the problem and algebra to show the general rule is true (4 marks);
describing the process to find the general rule (2 marks);
performing two further verifications and comment on how this means it is always true (1 mark).
Justifying is the mathematical process of reasoning why the general rule must always be true.
This is the most difficult part of the investigation to complete, but it is still possible to get some of the marks fairly easily.
This is marked against criteria J.
To get 1 mark you need to do two further verifications. If you used \(n=5\) for the verify step, then use \(n=6\) and \(n=7\) for further verifications. You should also state something like “As it works for many values it should work for all values”.
To get 2 marks you can usually explain the process to find the \(n\)th term:
for a linear sequence state what the common difference is and explain how you use the zeroth term to find the constant term;
for a quadratic sequence state the second difference and then \(an^2\). Then show the difference table to get the linear sequence and what the general term is for this linear sequence;
for a geometric sequence state what the common ratio is and the first term and how this gives the general rule through the formula.
To get the 3 or 4 marks, you need to show that the general rule is always true based on the context of the original problem. Usually this will involve some algebraic work from the diagrams, and some worded explanation alongside.
Example
Stage 1
Grey
4
Black
8
Total
12
Drag slider to grow pattern
You can use the interactive element to see how this pattern increases.
The key with these kinds of patterns is to look for the different shapes, here shown with different colours.
The grey square in the middle is always a square. The side length of the square is always one more than the stage number.
So the rule for the side length of the grey square is \(n+1\), and the rule for the area of the grey square is$$G=\left(n+1\right)^{2}$$
The number of black squares along each side is always the length of the grey square, which is one more than the stage number.
So the rule for the number of black squares on one side is \(n+1\), and since there are 4 sides, the total number of black squares is$$B=4\left(n+1\right)$$
The total area, \(A\), is the sum of \(G\) and \(B\).
So$$A=\left(n+1\right)^{2}+4\left(n+1\right)$$
If you found the quadratic sequence using the sequence method, then you will have a simplified version of this. In that case, you need to show that this simplifies to your general rule:$$\begin{align}A&=\left(n+1\right)^{2}+4\left(n+1\right)\\&=n^2+2n+1+4n+4\\&=n^2+6n+5\end{align}$$
Your Turn
Justify that the pattern below has general rule \(A=n^{2}+n\).
Stage 1
Grey
1
Black
1
Total
2
Drag slider to grow pattern
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The grey square in the middle is a square with side length \(n\). So $$G=n^{2}$$The black squares on the side form a row of width \(1\) and length \(n\). So$$B=n$$Since the area is the sum of these we have $$A=G+B=n^{2}+n$$
Your Turn
Justify that the pattern below has general rule \(A=n^{2}+n-1\).
Stage 1
Grey
1
Black
0
Total
1
Drag slider to grow pattern
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The grey square in the middle is a square with side length \(n\). So $$G=n^{2}$$The black squares on the bottom form a row of height \(1\) and the length is always one less than the square, so has length \(n-\). So$$B=n-1$$Since the area is the sum of these we have $$A=G+B=n^{2}+n-1$$
Your Turn
Justify that the pattern below has general rule \(A=n^{2}+2n\).
Stage 1
Grey
1
Black
2
Total
3
Drag slider to grow pattern
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The grey square in the middle is a square with side length \(n\). So $$G=n^{2}$$The black squares on the two sides each form a row of width \(1\) and length \(n\). As there are two rows$$B=2n$$Since the area is the sum of these we have $$A=G+B=n^{2}+2n$$
Your Turn
Justify that the pattern below has general rule \(A=n^{2}+2n+5\).
Stage 1
Grey
4
Black
4
Total
8
Drag slider to grow pattern
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The grey square in the middle is a square with side length \(n+1\) as it starts at \(2\). So $$G=\left(n+1\right)^{2}$$There are always \(4\) black squares on the corners. So$$B=4$$Since the area is the sum of these we have $$\begin{align}A&=G+B\\&=\left(n+1\right)^{2}+4\\&=n^{2}+2n+1+4\\&=n^{2}+2n+5\end{align}$$
Your Turn
Justify that the pattern below has general rule \(A=n^{2}+2n-1\).
Stage 1
Grey
1
Black
1
Red
0
Total
2
Drag slider to grow pattern
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The grey square in the middle is a square with side length \(n\). So $$G=n^{2}$$The row of black squares on the side has width \(1\) and length \(n\). So$$B=n$$The row of red squares on the bottom has height \(1\) and width \(n-1\), always one less than the square. So$$R=n-1$$
Since the area is the sum of these we have $$\begin{align}A&=G+B+R\\&=n^{2}+n+n-1\\&=n^{2}+2n-1\end{align}$$
Your Turn
Justify that the pattern below has general rule \(A=n^{2}+8n+20\).
Stage 1
Grey
9
Black
12
Red
8
Total
29
Drag slider to grow pattern
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The grey square in the middle is a square with side length \(n+2\) as it starts at \(3\). So $$G=\left(n+2\right)^{2}$$The rows of black squares on each side have width \(1\) and length \(n+2\). As there are \(4\) of them we have$$B=4\left(n+2\right)$$There are always \(8\) red squares near the corners. So$$R=8$$
Since the area is the sum of these we have $$\begin{align}A&=G+B+R\\&=\left(n+2\right)^{2}+4\left(n+2\right)+8\\&=n^{2}+4n+4+4n+8+8\\&=n^{2}+8n+20\end{align}$$
Your Turn
Justify that the pattern below has general rule \(A=4n^{2}+12n-3\).
Stage 1
Grey
9
Black
4
Total
13
Drag slider to grow pattern
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The side lengths of the grey square in the middle form the sequence$$3, 5, 7, 9, …$$which has general term \(2n+1\). Since these are the sides of the grey square, we have$$G=\left(2n+1\right)^{2}$$The length of the black rows on the \(4\) sides of the square form the sequence $$1,3,5,7,…$$which has general term \(2n-1\). So$$B=4\left(2n-1\right)$$Since the area is the sum of these we have $$\begin{align}A&=G+B\\&=\left(2n+1\right)^{2}+4\left(2n-1\right)\\&=4n^{2}+4n+1+8n-4\\&=4n^{2}+12n-3\end{align}$$
Your Turn
Justify that the pattern below has general rule \(A=4n^{2}+28n+25\).
Stage 1
Grey
25
Black
16
Red
12
Total
53
Drag slider to grow pattern
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The side lengths of the grey square in the middle form the sequence$$ 5, 7, 9, 11, …$$which has general term \(2n+3\). Since these are the sides of the grey square, we have$$G=\left(2n+3\right)^{2}$$The length of the black rows on the \(4\) sides of the square form the sequence $$5,7,9,11,…$$which has general term \(2n+3\). So$$B=4\left(2n+3\right)$$The length of the red rows on the \(4\) sides of the square form the sequence $$3,5,7,9,…$$which has general term \(2n+1\). So$$R=4\left(2n+1\right)$$Since the area is the sum of these we have $$\begin{align}A&=G+B+R\\&=\left(2n+3\right)^{2}+4\left(2n+3\right)+4\left(2n+1\right)\\&=4n^{2}+12n+9+8n+12+8n+4\\&=4n^{2}+28n+25\end{align}$$
Your Turn
Justify that the pattern below has general rule \(A=4n+4\).
Stage 1
Total Units
8
Inner Gap
1
Drag slider to grow pattern
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The side lengths of the outer grey square form the sequence \(n+2\) as they start at \(3\). So$$G=\left(n+2\right)^{2}$$The side length of the square hole form the sequence \(n\). So$$H=n^{2}$$Since the area is the difference of these we have $$\begin{align}A&=G-H\\&=\left(n+2\right)^{2}-n^{2}\\&=n^{2}+4n+4-n^{2}\\&=4n+4\end{align}$$
Your Turn
Justify that the pattern below has general rule \(A=5n^{2}+4n+4\).
Stage 1
Grey
9
Black
4
Total
13
Drag slider to grow pattern
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The side lengths of the grey square in the middle form the sequence \(n+2\). So$$G=\left(n+2\right)^{2}$$The side lengths of the black squares on the \(4\) sides of the grey square form the sequence \(n\). So$$B=4n^{2}$$Since the area is the sum of these we have $$\begin{align}A&=G+B\\&=\left(n+2\right)^{2}+4n^{2}\&=n^{2}+4n+4+4n^{2}\\&=5n^{2}+4n+4\end{align}$$
Your Turn
Justify that the pattern below has general rule \(A=5n^2+6n+5\).
Stage 1
Total Units
16
Drag slider to grow pattern
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The side lengths of the grey square in the middle form the sequence \(n+3\) since they start at \(4\). But the \(4\) corners are always removed. So$$G=\left(n+3\right)^{2}-4$$The side lengths of the black squares on the \(4\) corners of the grey square form the sequence \(n\). So$$B=4n^{2}$$Since the area is the sum of these we have $$\begin{align}A&=G+B\\&=\left(n+3\right)^{2}-4+4n^{2}\\&=n^{2}+6n+9-4+4n^{2}\\&=5n^{2}+6n+5\end{align}$$
Your Turn
Justify that the pattern below has general rule \(A=4n^{2}+4n+9\).
Stage 1
Total Units
17
Drag slider to grow pattern
-
The side lengths of the outer grey square form the sequence$$ 5, 7, 9, 11, …$$which has general term \(2n+3\). Since these are the sides of the grey square (assuming completely filled), we have$$G=\left(2n+3\right)^{2}$$The hole inside the grey square has side lengths which form the sequence $$3,5,7,9,…$$which has general term \(2n+1\). So the square hole has general term$$H=\left(2n+1\right)^{2}$$The side lengths of the inner black square form the sequence$$ 1,3,5,7, …$$which has general term \(2n-1\). Since these are the sides of the black square , we have$$B=\left(2n-1\right)^{2}$$Since the area is made up of these we have $$\begin{align}A&=G-H+B\\&=\left(2n+3\right)^{2}-\left(2n+1\right)^{2}+\left(2n-1\right)^{2}\\&=4n^{2}+12n+9\\&\quad-\left(4n^{2}+4n+1\right)\\&\quad+4n^{2}-4n+1\\&=4n^{2}+4n+9\end{align}$$