Expressions Using Letter-Numbers | FIO

Question 23

Observe the pattern below. How many squares will be there in Step 4, Step 10, Step 50? Write a general formula. How would the formula change if we want to count the number of vertices of all the squares?

Solution

We will look for a pattern in the number of squares and vertices. Then we will write a general rule for them.

Step 1 — Count squares in each step

Let us count the squares in each pattern. In Step 1, there is 1 square in the middle. There are 4 squares in the arms. The total number of squares is: $1 + 4$ $= 5$

$\boxed{\text{5 squares}}$ In Step 2, there is 1 square in the middle. There are 4 arms. Each arm has 2 squares. The total number of squares is: $1 + (4 \times 2)$ $= 1 + 8$ $= 9$ $\boxed{\text{9 squares}}$ In Step 3, there is 1 square in the middle. There are 4 arms. Each arm has 3 squares. The total number of squares is: $1 + (4 \times 3)$ $= 1 + 12$ $= 13$ $\boxed{\text{13 squares}}$

Diagram 2

Step 2 — Find the general formula for squares

Let $n$ be the step number. Let $S(n)$ be the number of squares in step $n$ . We see that $S(1) = 5$ , $S(2) = 9$ , $S(3) = 13$ . The number of squares increases by 4 each time. This is an arithmetic pattern. Each arm has $n$ squares. There are 4 arms. There is always 1 central square. So, the formula for squares is: $S(n) = 1 + (4 \times n)$

$\boxed{S(n) = 4n + 1}$

Step 3 — Calculate squares for specific steps

We use the formula $S(n) = 4n + 1$ . For Step 4, we put $n = 4$ . $S(4) = (4 \times 4) + 1$ $= 16 + 1$ $= 17$

$\boxed{\text{17 squares}}$ For Step 10, we put $n = 10$ . $S(10) = (4 \times 10) + 1$ $= 40 + 1$ $= 41$ $\boxed{\text{41 squares}}$ For Step 50, we put $n = 50$ . $S(50) = (4 \times 50) + 1$ $= 200 + 1$ $= 201$ $\boxed{\text{201 squares}}$

Step 4 — Find the general formula for vertices

Let us count the distinct vertices in each pattern. In Step 1, the central square has 4 vertices. Each of the 4 arms has 1 square. Each square in an arm adds 2 new vertices. So, each arm adds $1 \times 2 = 2$ new vertices. The total number of vertices is: $4 + (4 \times 2)$ $= 4 + 8$ $= 12$

$\boxed{\text{12 vertices}}$ In Step 2, the central square has 4 vertices. Each of the 4 arms has 2 squares. Each square in an arm adds 2 new vertices. So, each arm adds $2 \times 2 = 4$ new vertices. The total number of vertices is: $4 + (4 \times 4)$ $= 4 + 16$ $= 20$ $\boxed{\text{20 vertices}}$ In Step 3, the central square has 4 vertices. Each of the 4 arms has 3 squares. Each square in an arm adds 2 new vertices. So, each arm adds $3 \times 2 = 6$ new vertices. The total number of vertices is: $4 + (4 \times 6)$ $= 4 + 24$ $= 28$ $\boxed{\text{28 vertices}}$ Let $V(n)$ be the number of vertices in step $n$ . We see that $V(1) = 12$ , $V(2) = 20$ , $V(3) = 28$ . The number of vertices increases by 8 each time. This is an arithmetic pattern. Each arm has $n$ squares. Each square adds 2 new vertices. So, each arm adds $n \times 2$ new vertices. There are 4 arms. There are always 4 vertices from the central square. So, the formula for vertices is: $V(n) = 4 + (4 \times 2 \times n)$ $\boxed{V(n) = 8n + 4}$

<DIAGRAM: Three patterns of orange squares. The first pattern is a cross shape with 5 squares. The second pattern is a larger cross with 9 squares. The third pattern is an even larger cross with 13 squares.>

Step 5 — How the formula changes for vertices

The formula for the number of squares is $S(n) = 4n + 1$ . The formula for the number of vertices is $V(n) = 8n + 4$ . We can write the vertices formula using the squares formula. Let us rewrite $V(n)$ . $V(n) = 8n + 4$ We can take out a common factor of 2. $V(n) = 2 \times (4n + 2)$ We know $S(n) = 4n + 1$ . So, we can write $4n + 2$ as $(4n + 1) + 1$ . $V(n) = 2 \times ((4n + 1) + 1)$ $V(n) = 2 \times (S(n) + 1)$ $V(n) = 2S(n) + 2$ The formula for vertices is twice the number of squares plus 2.

Answer

(i) Step 4: 17 squares, Step 10: 41 squares, Step 50: 201 squares. (ii) General formula for squares: $S(n) = 4n + 1$ . (iii) The formula for vertices would be $V(n) = 8n + 4$ . This formula is $V(n) = 2S(n) + 2$ , which means it is twice the number of squares plus 2.

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Q10

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(a) $30x + 20y$

(b) $(30 + 20) \times (x + y)$

(d) $(30 + 20) \times x + y$

(e) $30x - 20y$

Q11

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(a) p + q + r

(b) p + q + 2r

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Q12

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(b) What can we say about the snail’s movement if d > u?

Q13

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Q14

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Q15

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Q16

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Q17

Add the expressions given below:

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Q18

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(b) $-15x + 13 - 9y$ from $7y - 10 + 3x$

(d) $9a - 6b + 14$ from $6a - (9b + 18)$

(e) $10x + 2 + 10y$ from $-3y + 8 - 3x$

(f) $8g + 4h - 10$ from $7h - 8g + 20$

Q19

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Q20

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Q21

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Q22

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Q23