We Distribute, Yet Things Multiply

Given: The middle number of a 3×3 frame in the multiplication grid is pq, meaning the row header is p and the column header is q.

In the multiplication grid, each entry is the product of its row number and column number. If the centre cell corresponds to row p and column q, then:
- The row above has row number (p−1), and the row below has row number (p+1).
- The column to the left has column number (q−1), and the column to the right has column number (q+1).

Therefore, the nine entries of the 3×3 frame are:

$$\begin{array}{|c|c|c|}\hline (p-1)(q-1) & (p-1)q & (p-1)(q+1) \\ \hline p(q-1) & pq & p(q+1) \\ \hline (p+1)(q-1) & (p+1)q & (p+1)(q+1) \\ \hline\end{array}$$

Expanding each:
- Top-left: $pq - p - q + 1$
- Top-centre: $pq - q$
- Top-right: $pq - p + q - 1$ (wait, let us keep them in factored form as that is cleaner)

The expressions in factored form are:

| $(p-1)(q-1)$ | $(p-1)q$ | $(p-1)(q+1)$ |
|---|---|---|
| $p(q-1)$ | $pq$ | $p(q+1)$ |
| $(p+1)(q-1)$ | $(p+1)q$ | $(p+1)(q+1)$ |

For example, using the 3×3 frame with centre 4×6 = 24 (p=4, q=6):
- The frame entries are: 3×5=15, 3×6=18, 3×7=21, 4×5=20, 4×6=24, 4×7=28, 5×5=25, 5×6=30, 5×7=35, which matches the grid.

Using the distributive property: $(a+b)(c+d) = ac + ad + bc + bd$.

(i) $(3 + u)(v - 3)$
$$= 3 \cdot v + 3 \cdot (-3) + u \cdot v + u \cdot (-3)$$
$$= 3v - 9 + uv - 3u$$
$$\boxed{= uv + 3v - 3u - 9}$$

(ii) $\dfrac{2}{3}(15 + 6a)$
$$= \frac{2}{3} \times 15 + \frac{2}{3} \times 6a$$
$$= 10 + 4a$$
$$\boxed{= 4a + 10}$$

(iii) $(10a + b)(10c + d)$
$$= 10a \cdot 10c + 10a \cdot d + b \cdot 10c + b \cdot d$$
$$\boxed{= 100ac + 10ad + 10bc + bd}$$

(iv) $(3 - x)(x - 6)$
$$= 3 \cdot x + 3 \cdot (-6) + (-x) \cdot x + (-x)(-6)$$
$$= 3x - 18 - x^2 + 6x$$
$$\boxed{= -x^2 + 9x - 18}$$

(v) $(-5a + b)(c + d)$
$$= -5a \cdot c + (-5a) \cdot d + b \cdot c + b \cdot d$$
$$\boxed{= -5ac - 5ad + bc + bd}$$

(vi) $(5 + z)(y + 9)$
$$= 5 \cdot y + 5 \cdot 9 + z \cdot y + z \cdot 9$$
$$= 5y + 45 + zy + 9z$$
$$\boxed{= zy + 5y + 9z + 45}$$

We need pairs $(a, b)$ and $(a+2, b-4)$ such that $a \times b = (a+2)(b-4)$.

Expanding the right side:
$$(a+2)(b-4) = ab - 4a + 2b - 8$$

For the products to be equal:
$$ab = ab - 4a + 2b - 8$$
$$0 = -4a + 2b - 8$$
$$4a - 2b + 8 = 0$$
$$2a - b + 4 = 0$$
$$b = 2a + 4$$

So any pair of the form $(a,\ 2a+4)$ works.

Example 1: Let $a = 1$, then $b = 2(1)+4 = 6$.
- $1 \times 6 = 6$
- $(1+2)(6-4) = 3 \times 2 = 6$ ✓

Example 2: Let $a = 3$, then $b = 2(3)+4 = 10$.
- $3 \times 10 = 30$
- $(3+2)(10-4) = 5 \times 6 = 30$ ✓

Example 3: Let $a = 5$, then $b = 2(5)+4 = 14$.
- $5 \times 14 = 70$
- $(5+2)(14-4) = 7 \times 10 = 70$ ✓

Using the distributive property, multiply each term of the first expression by each term of the second.

(i) $(a + ab - 3b^2)(4 + b)$

$$= a(4 + b) + ab(4 + b) + (-3b^2)(4 + b)$$
$$= 4a + ab + 4ab + ab^2 - 12b^2 - 3b^3$$
$$= 4a + (ab + 4ab) + ab^2 - 12b^2 - 3b^3$$
$$\boxed{= 4a + 5ab + ab^2 - 12b^2 - 3b^3}$$

(ii) $(4y + 7)(y + 11z - 3)$

$$= 4y(y + 11z - 3) + 7(y + 11z - 3)$$
$$= 4y^2 + 44yz - 12y + 7y + 77z - 21$$
$$= 4y^2 + 44yz + (-12y + 7y) + 77z - 21$$
$$\boxed{= 4y^2 + 44yz - 5y + 77z - 21}$$

(i) $(a - b)(a + b)$
$$= a \cdot a + a \cdot b - b \cdot a - b \cdot b$$
$$= a^2 + ab - ab - b^2$$
$$\boxed{= a^2 - b^2}$$

(ii) $(a - b)(a^2 + ab + b^2)$
$$= a(a^2 + ab + b^2) - b(a^2 + ab + b^2)$$
$$= a^3 + a^2b + ab^2 - a^2b - ab^2 - b^3$$
$$= a^3 + (a^2b - a^2b) + (ab^2 - ab^2) - b^3$$
$$\boxed{= a^3 - b^3}$$

(iii) $(a - b)(a^3 + a^2b + ab^2 + b^3)$
$$= a(a^3 + a^2b + ab^2 + b^3) - b(a^3 + a^2b + ab^2 + b^3)$$
$$= a^4 + a^3b + a^2b^2 + ab^3 - a^3b - a^2b^2 - ab^3 - b^4$$
$$= a^4 + (a^3b - a^3b) + (a^2b^2 - a^2b^2) + (ab^3 - ab^3) - b^4$$
$$\boxed{= a^4 - b^4}$$

Pattern observed:
$$(a-b)(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \cdots + b^{n-1}) = a^n - b^n$$

The next identity in the pattern (n = 5) would be:
$$(a - b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4) = a^5 - b^5$$

Verification:
$$a(a^4 + a^3b + a^2b^2 + ab^3 + b^4) - b(a^4 + a^3b + a^2b^2 + ab^3 + b^4)$$
$$= a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 - a^4b - a^3b^2 - a^2b^3 - ab^4 - b^5$$
$$= a^5 - b^5 \checkmark$$

Given: $(a-b)^2$ and $(b-a)^2$.

Key observation: Note that $(b - a) = -(a - b)$.

Therefore:
$$(b - a)^2 = (-(a-b))^2 = (-1)^2 \cdot (a-b)^2 = (a-b)^2$$

Conclusion: $(a-b)^2 = (b-a)^2$. Neither is greater; they are always equal to each other for all values of $a$ and $b$.

We use the identity $(a+b)(a-b) = a^2 - b^2$.

We need $a^2 - b^2 = 100$, i.e., $(a+b)(a-b) = 100$.

Method: Choose $a + b$ and $a - b$ as factor pairs of 100.

Let $a + b = 50$ and $a - b = 2$.
Adding: $2a = 52 \Rightarrow a = 26$, $b = 24$.

$$26^2 - 24^2 = 676 - 576 = 100 \checkmark$$

Another way: Let $a + b = 20$, $a - b = 5$ — but this gives non-integers. Let $a+b = 25$, $a-b = 4$: $a = 14.5$ (not integer).

A simple integer answer: $\boxed{26^2 - 24^2 = 100}$.

Alternatively, $a+b=10, a-b=10 \Rightarrow a=10, b=0$: $10^2 - 0^2 = 100$. ✓

So $100 = 10^2 - 0^2$ or $100 = 26^2 - 24^2$.

We use the identity $a^2 = (a+b)(a-b) + b^2$, choosing $b$ to make the multiplication easy, or we use $(a+b)^2 = a^2 + 2ab + b^2$ or $(a-b)^2 = a^2 - 2ab + b^2$.

$406^2$: Use $(a+b)^2$ with $a = 400$, $b = 6$.
$$406^2 = (400 + 6)^2 = 400^2 + 2(400)(6) + 6^2 = 160000 + 4800 + 36 = \boxed{164836}$$

$72^2$: Use $(a-b)^2$ with $a = 70$, $b = -2$, i.e., $(70+2)^2$.
$$72^2 = (70 + 2)^2 = 70^2 + 2(70)(2) + 2^2 = 4900 + 280 + 4 = \boxed{5184}$$

$145^2$: Use $(a+b)^2$ with $a = 140$, $b = 5$, or use $a = 150$, $b = 5$: $(150-5)^2$.
$$145^2 = (150 - 5)^2 = 150^2 - 2(150)(5) + 5^2 = 22500 - 1500 + 25 = \boxed{21025}$$

$1097^2$: Use $(a-b)^2$ with $a = 1100$, $b = 3$.
$$1097^2 = (1100 - 3)^2 = 1100^2 - 2(1100)(3) + 3^2 = 1210000 - 6600 + 9 = \boxed{1203409}$$

$124^2$: Use $(a+b)^2$ with $a = 120$, $b = 4$.
$$124^2 = (120 + 4)^2 = 120^2 + 2(120)(4) + 4^2 = 14400 + 960 + 16 = \boxed{15376}$$

Pattern 1 refers to $(a+b)^2 = a^2 + 2ab + b^2$ and Pattern 2 refers to $(a-b)^2 = a^2 - 2ab + b^2$ (the square identities derived from the distributive property).

These identities are derived purely using the distributive property of multiplication over addition, which holds for all real numbers — including negative integers and fractions.

For negative integers: Let $a = -3$, $b = -2$.
$$(-3 + (-2))^2 = (-5)^2 = 25$$
$$(-3)^2 + 2(-3)(-2) + (-2)^2 = 9 + 12 + 4 = 25 \checkmark$$

For fractions: Let $a = \dfrac{1}{2}$, $b = \dfrac{1}{3}$.
$$\left(\frac{1}{2} + \frac{1}{3}\right)^2 = \left(\frac{5}{6}\right)^2 = \frac{25}{36}$$
$$\left(\frac{1}{2}\right)^2 + 2 \cdot \frac{1}{2} \cdot \frac{1}{3} + \left(\frac{1}{3}\right)^2 = \frac{1}{4} + \frac{1}{3} + \frac{1}{9} = \frac{9}{36} + \frac{12}{36} + \frac{4}{36} = \frac{25}{36} \checkmark$$

Conclusion: Both patterns hold for all real numbers — counting numbers, negative integers, and fractions — because they are algebraic identities based on the distributive property.

Identity 1A: $(a+b)^2 = a^2 + 2ab + b^2$
Identity 1B: $(a-b)^2 = a^2 - 2ab + b^2$
Identity 1C: $(a+b)(a-b) = a^2 - b^2$

(i) $46^2$ using $(a+b)^2$:
Write $46 = 40 + 6$, so $a = 40$, $b = 6$.
$$46^2 = (40+6)^2 = 40^2 + 2(40)(6) + 6^2 = 1600 + 480 + 36 = \boxed{2116}$$

(ii) $397 \times 403$ using $(a+b)(a-b)$:
Write $397 = 400 - 3$ and $403 = 400 + 3$, so $a = 400$, $b = 3$.
$$397 \times 403 = (400-3)(400+3) = 400^2 - 3^2 = 160000 - 9 = \boxed{159991}$$

(iii) $91^2$ using $(a-b)^2$:
Write $91 = 100 - 9$, so $a = 100$, $b = 9$.
$$91^2 = (100-9)^2 = 100^2 - 2(100)(9) + 9^2 = 10000 - 1800 + 81 = \boxed{8281}$$

(iv) $43 \times 45$ using $(a+b)(a-b)$:
Write $43 = 44 - 1$ and $45 = 44 + 1$, so $a = 44$, $b = 1$.
$$43 \times 45 = (44-1)(44+1) = 44^2 - 1^2 = 1936 - 1 = \boxed{1935}$$

(i) $(p-1)(p+11)$
Using distributive property:
$$= p(p+11) - 1(p+11) = p^2 + 11p - p - 11 = \boxed{p^2 + 10p - 11}$$

(ii) $(3a - 9b)(3a + 9b)$
Using Identity 1C: $(x-y)(x+y) = x^2 - y^2$ with $x = 3a$, $y = 9b$:
$$= (3a)^2 - (9b)^2 = 9a^2 - 81b^2 = \boxed{9a^2 - 81b^2}$$

(iii) $-(2y+5)(3y+4)$
First expand $(2y+5)(3y+4)$:
$$= 2y(3y+4) + 5(3y+4) = 6y^2 + 8y + 15y + 20 = 6y^2 + 23y + 20$$
Now multiply by $-1$:
$$\boxed{= -6y^2 - 23y - 20}$$

(iv) $(6x + 5y)^2$
Using Identity 1A: $(a+b)^2 = a^2 + 2ab + b^2$ with $a = 6x$, $b = 5y$:
$$= (6x)^2 + 2(6x)(5y) + (5y)^2 = 36x^2 + 60xy + 25y^2$$
$$\boxed{= 36x^2 + 60xy + 25y^2}$$

(v) $\left(2x - \dfrac{1}{2}\right)^2$
Using Identity 1B: $(a-b)^2 = a^2 - 2ab + b^2$ with $a = 2x$, $b = \dfrac{1}{2}$:
$$= (2x)^2 - 2(2x)\left(\frac{1}{2}\right) + \left(\frac{1}{2}\right)^2 = 4x^2 - 2x + \frac{1}{4}$$
$$\boxed{= 4x^2 - 2x + \frac{1}{4}}$$

(vi) $(7p) \times (3r) \times (p+2)$
First multiply the constants and variables:
$$7p \times 3r = 21pr$$
Now:
$$21pr \times (p+2) = 21pr \cdot p + 21pr \cdot 2 = 21p^2r + 42pr$$
$$\boxed{= 21p^2r + 42pr}$$

(i) Two more than a square number:

'A square number' means $s^2$ (the square of some number $s$). 'Two more than' means we add 2.

So the expression is $s^2 + 2$.

Also, $2 + s$ is not a square number plus 2 (it's just $s$ plus 2). $(s+2)^2$ means the square of $(s+2)$, not a square plus 2. $s^2 + 4$ would be four more than a square. $2s^2$ is twice a square. $2^2 s = 4s$ is not a square plus 2.

$$\boxed{s^2 + 2}$$

(ii) The sum of the squares of two consecutive numbers:

Two consecutive numbers can be written as $m$ and $m+1$. Their squares are $m^2$ and $(m+1)^2$. The sum is:
$$m^2 + (m+1)^2$$

Also, $m^2 + (m-1)^2$ represents the sum of squares of $m$ and $m-1$, which are also consecutive numbers. So this is also valid.

And $(2m)^2 + (2m+1)^2$ represents consecutive numbers $2m$ and $2m+1$ (consecutive even and odd), so this is also a valid specific case.

$$\boxed{m^2 + (m+1)^2, \quad m^2 + (m-1)^2, \quad (2m)^2 + (2m+1)^2}$$

Note: $m^2 + n^2$ is the sum of squares of any two numbers, not necessarily consecutive. $(m+n)^2 \neq m^2 + n^2$ in general. $m^2 + 1$ is only valid if the consecutive number is 1, which is too restrictive.

Observation: In every 2×2 square from a calendar, the product of one diagonal is always 7 less than the product of the other diagonal.

Explanation using algebra:

Label the top-left number of any 2×2 square as $a$. Since calendar dates in the same row differ by 1, and dates in the next row differ by 7:

$$\begin{array}{|c|c|}\hline a & a+1 \\ \hline a+7 & a+8 \\ \hline\end{array}$$

Diagonal 1 (top-left × bottom-right):
$$a(a+8) = a^2 + 8a$$

Diagonal 2 (top-right × bottom-left):
$$(a+1)(a+7) = a^2 + 7a + a + 7 = a^2 + 8a + 7$$

Difference:
$$(a+1)(a+7) - a(a+8) = (a^2 + 8a + 7) - (a^2 + 8a) = 7$$

Conclusion: The product of the top-right and bottom-left numbers is always 7 more than the product of the top-left and bottom-right numbers. This is because consecutive dates differ by 1 (horizontally) and 7 (vertically), and the algebraic identity shows the difference is always exactly 7, regardless of which 2×2 square is chosen.

(i) $(k+1)(k+2) - (k+3)$ is always 2.

Expand:
$$(k+1)(k+2) = k^2 + 2k + k + 2 = k^2 + 3k + 2$$
$$k^2 + 3k + 2 - (k+3) = k^2 + 3k + 2 - k - 3 = k^2 + 2k - 1$$

For $k = 1$: $1 + 2 - 1 = 2$. For $k = 2$: $4 + 4 - 1 = 7 \neq 2$.

This statement is FALSE. $(k+1)(k+2) - (k+3) = k^2 + 2k - 1$, which is not always 2.

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(ii) $(2q+1)(2q-3)$ is a multiple of 4.

Expand:
$$(2q+1)(2q-3) = 4q^2 - 6q + 2q - 3 = 4q^2 - 4q - 3$$
$$= 4(q^2 - q) - 3$$

$4(q^2 - q)$ is a multiple of 4, but subtracting 3 gives a number that is $\equiv 1 \pmod{4}$ (since $-3 \equiv 1 \pmod 4$).

For $q = 1$: $(3)(-1) = -3$, not a multiple of 4.

This statement is FALSE.

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(iii) Squares of even numbers are multiples of 4; squares of odd numbers are 1 more than multiples of 8.

Even numbers: Let $n = 2m$.
$$(2m)^2 = 4m^2$$
This is always a multiple of 4. ✓

Odd numbers: Let $n = 2m+1$.
$$(2m+1)^2 = 4m^2 + 4m + 1 = 4m(m+1) + 1$$
Now, $m(m+1)$ is the product of two consecutive integers, so it is always even. Let $m(m+1) = 2t$.
$$= 4(2t) + 1 = 8t + 1$$
This is always 1 more than a multiple of 8. ✓

This statement is TRUE.

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(iv) $(6n+2)^2 - (4n+3)^2$ is 5 less than a square number.

Using identity $(a^2 - b^2) = (a+b)(a-b)$ with $a = 6n+2$, $b = 4n+3$:
$$a + b = (6n+2) + (4n+3) = 10n + 5$$
$$a - b = (6n+2) - (4n+3) = 2n - 1$$
$$(6n+2)^2 - (4n+3)^2 = (10n+5)(2n-1)$$

Expand:
$$= 20n^2 - 10n + 10n - 5 = 20n^2 - 5$$

Now check if this is 5 less than a square:
$$20n^2 - 5 + 5 = 20n^2$$

Is $20n^2$ a perfect square? $20n^2 = 4 \cdot 5 \cdot n^2 = (2n)^2 \cdot 5$. This is a perfect square only when $5$ is a perfect square, which it is not (in general).

For $n = 1$: $20(1) - 5 = 15$. Is $15 + 5 = 20$ a perfect square? No.

This statement is FALSE.

Given:
Let the first number be $a = 7m + 3$ for some integer $m$.
Let the second number be $b = 7n + 5$ for some integer $n$.

Sum:
$$a + b = (7m + 3) + (7n + 5) = 7(m + n) + 8 = 7(m + n + 1) + 1$$
Remainder when sum is divided by 7 = $\boxed{1}$.

Difference:
$$a - b = (7m + 3) - (7n + 5) = 7(m - n) - 2 = 7(m - n - 1) + 5$$
Remainder when difference is divided by 7 = $\boxed{5}$.

(Note: If b > a, then $b - a = 7(n-m) + 2$, giving remainder 2.)

Product:
$$a \times b = (7m + 3)(7n + 5)$$
$$= 49mn + 35m + 21n + 15$$
$$= 7(7mn + 5m + 3n + 2) + 1$$
Remainder when product is divided by 7 = $\boxed{1}$.

(Since $3 \times 5 = 15 = 2 \times 7 + 1$, the remainder is 1.)

Exploration:
- Consecutive numbers 4, 5, 6: $5^2 - 4 \times 6 = 25 - 24 = 1$
- Consecutive numbers 7, 8, 9: $8^2 - 7 \times 9 = 64 - 63 = 1$
- Consecutive numbers 10, 11, 12: $11^2 - 10 \times 12 = 121 - 120 = 1$

Pattern: The result is always 1.

Algebraic equation:
Let the three consecutive numbers be $(n-1)$, $n$, $(n+1)$.
$$n^2 - (n-1)(n+1) = 1$$

Verification by expansion:

Left side:
$$n^2 - (n-1)(n+1) = n^2 - (n^2 - 1) = n^2 - n^2 + 1 = 1$$

Right side: $1$.

LHS = RHS = 1. ✓

This is a true identity: $n^2 - (n-1)(n+1) = 1$ for all values of $n$.

Let the two numbers be $a$ and $b$.

Step 1: Add the two numbers: $a + b$.

Step 2: Half of the sum of the two numbers: $\dfrac{a+b}{2}$.

Step 3: Multiply: $(a + b) \times \dfrac{a+b}{2}$.

Algebraic expression:
$$\frac{(a+b)(a+b)}{2} = \frac{(a+b)^2}{2}$$

Proof:

We need to show that $(a+b) \times \dfrac{a+b}{2} = \dfrac{(a+b)^2}{2}$.

$$\text{LHS} = (a+b) \times \frac{(a+b)}{2} = \frac{(a+b)^2}{2} = \text{RHS}$$

This is exactly half of the square of the sum of the two numbers $(a+b)^2$.

Hence proved. The result of multiplying the sum of two numbers by half their sum equals half the square of their sum.

We use the identity $(a+b)(a-b) = a^2 - b^2$, which tells us that for a fixed sum, the product is maximised when the two numbers are equal (or as close as possible).

(i) $14 \times 26$ or $16 \times 24$:

Note that $14 + 26 = 40$ and $16 + 24 = 40$. Both pairs have the same sum.

Write each as $(20 - d)(20 + d) = 400 - d^2$:
- $14 \times 26 = (20-6)(20+6) = 400 - 36 = 364$
- $16 \times 24 = (20-4)(20+4) = 400 - 16 = 384$

Since 16 < 36, we have 400 - 16 > 400 - 36.

\boxed{16 \times 24 > 14 \times 26}

(ii) $25 \times 75$ or $26 \times 74$:

Note that $25 + 75 = 100$ and $26 + 74 = 100$. Both pairs have the same sum.

Write each as $(50 - d)(50 + d) = 2500 - d^2$:
- $25 \times 75 = (50-25)(50+25) = 2500 - 625 = 1875$
- $26 \times 74 = (50-24)(50+24) = 2500 - 576 = 1924$

Since 576 < 625, we have 2500 - 576 > 2500 - 625.

\boxed{26 \times 74 > 25 \times 75}

Note: The figure is not visible, but based on the description, we assume the park consists of a large square (or rectangle) with two smaller square plots of side $g$ ft. each embedded in it, with a walking path of width $w$ ft. surrounding them.

Assumption (standard version of this problem): The overall park is a rectangle. The two green square plots each have side $g$ ft. (area $g^2$ sq. ft. each). The walking path of width $w$ ft. surrounds the green plots.

If the two green squares are placed side by side, the total green area = $2g^2$ sq. ft.

The overall dimensions of the park (including the path) would be:
- Length: $2g + 3w$ (path on both ends and between the squares)
- Width: $g + 2w$ (path on top and bottom)

Total area of park:
$$(2g + 3w)(g + 2w) = 2g^2 + 4gw + 3gw + 6w^2 = 2g^2 + 7gw + 6w^2$$

Area to be tiled (walking path) = Total area − Green area:
$$= (2g^2 + 7gw + 6w^2) - 2g^2$$
$$\boxed{= 7gw + 6w^2}$$

(The exact expression depends on the figure; the method above illustrates the approach.)

Note: The figure is not visible in the OCR. Based on standard NCERT Class 8 pattern questions of this type, we assume the pattern involves figures where:
- Step 1 has 1 unit
- Step 2 has 3 units (or follows a specific rule)

For a common pattern where each step adds a row/column:

Pattern A (L-shaped or cross pattern): Step $n$ has $n^2$ units or $2n-1$ units.

For the most common version where Step $n$ has $n^2$ basic units:

(i) The next figure (Step 4) would be a $4 \times 4$ arrangement = 16 units.

(ii) Step 10 would have $10^2 = 100$ basic units.

(iii) Expression for Step $y$: $y^2$ basic units.

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For a pattern where Step $n$ has $2n + 1$ units (another common type):

(i) Draw the next figure by adding 2 more units.

(ii) Step 10: $2(10) + 1 = 21$ basic units.

(iii) Expression for Step $y$: $2y + 1$ basic units.

*Please refer to the actual figure in the textbook to determine the exact pattern and apply the method above accordingly.*