Patterns in Mathematics

Given: Various number sequences represented pictorially in Table 2.

Concept: Each sequence follows a specific pattern; to draw the next picture, we identify the rule and extend it.

Working and next pictures for each sequence:

1. Counting Numbers (1, 2, 3, 4, 5, ...): Each picture has one more dot than the previous. The next picture after 5 dots is 6 dots arranged in a row.

2. Odd Numbers (1, 3, 5, 7, 9, ...): Each picture adds 2 more dots. The next picture after 9 dots is 11 dots.

3. Even Numbers (2, 4, 6, 8, 10, ...): Each picture adds 2 more dots. The next picture after 10 dots is 12 dots.

4. Triangular Numbers (1, 3, 6, 10, 15, ...): Each picture forms a triangle by adding one more row. The next triangular number is $15 + 6 = 21$, so the next picture is a triangle with 6 rows (21 dots total).

5. Square Numbers (1, 4, 9, 16, 25, ...): Each picture is a square grid. The next square after $5 \times 5 = 25$ is a **$6 \times 6$ grid (36 dots).

6. Cube Numbers (1, 8, 27, 64, 125, ...):** Each picture is a cube. The next cube after $5^3 = 125$ is a **$6 \times 6 \times 6$ cube (216 dots).

7. Virahānka Numbers (1, 1, 2, 3, 5, 8, 13, ...):** Each number is the sum of the two preceding numbers. The next number is $8 + 13 = \mathbf{21}$.

8. Powers of 2 (1, 2, 4, 8, 16, ...): Each picture doubles. The next picture after 16 is 32.

Answer: Draw each sequence as described above, extending by one step following the identified rule.

Triangular Numbers (1, 3, 6, 10, 15, ...):

These are called triangular numbers because the dots representing each number can be arranged in the shape of an equilateral triangle.
- 1 dot → a triangle with 1 row
- 3 dots → a triangle with 2 rows (1 + 2)
- 6 dots → a triangle with 3 rows (1 + 2 + 3)
- 10 dots → a triangle with 4 rows (1 + 2 + 3 + 4)
- 15 dots → a triangle with 5 rows (1 + 2 + 3 + 4 + 5)

In general, the $n$-th triangular number $= 1 + 2 + 3 + \cdots + n = \dfrac{n(n+1)}{2}$.

Square Numbers (1, 4, 9, 16, 25, ...):

These are called square numbers because the dots can be arranged in a perfect square grid.
- $1 = 1 \times 1$, $4 = 2 \times 2$, $9 = 3 \times 3$, $16 = 4 \times 4$, $25 = 5 \times 5$

In general, the $n$-th square number $= n^2$. The shape formed is always a square.

Cube Numbers (1, 8, 27, 64, 125, ...):

These are called cubes because the dots can be arranged to fill a perfect cube (3D box with equal sides).
- $1 = 1^3$, $8 = 2^3$, $27 = 3^3$, $64 = 4^3$, $125 = 5^3$

In general, the $n$-th cube number $= n^3$. The shape formed is always a cube.

Answer: The names come from the geometric shapes that the respective numbers of dots can form — triangles, squares, and cubes.

Given: 36 is both a triangular number and a square number.

Verification:
- As a triangular number: $36 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = \dfrac{8 \times 9}{2} = 36$. So 36 dots can be arranged in a triangle with 8 rows.
- As a square number: $36 = 6 \times 6 = 6^2$. So 36 dots can be arranged in a $6 \times 6$ square grid.

Picture description:
- Draw a triangle with 8 rows: row 1 has 1 dot, row 2 has 2 dots, ..., row 8 has 8 dots. Total = 36 dots.
- Draw a $6 \times 6$ square grid with 36 dots.

Other examples of numbers with multiple representations:
- 1 is a triangular number ($T_1$), a square number ($1^2$), and a cube ($1^3$).
- 9 is a square number ($3^2$) and can also be seen as the sum of first 3 odd numbers ($1+3+5=9$).
- 10 is a triangular number ($T_4$) and also the sum of first 4 counting numbers.

Answer: 36 dots form both a triangle (8 rows) and a $6 \times 6$ square, showing the same number can have different geometric representations.

Given: The sequence shown in the image represents hexagonal numbers: $1, 7, 19, 37, 61, \ldots$

Name: These are called Hexagonal Numbers because the dots can be arranged in the shape of a hexagon (a six-sided figure).

Pattern:
- $1^{\text{st}}$ hexagonal number: $1$
- $2^{\text{nd}}$ hexagonal number: $7$ (difference = 6)
- $3^{\text{rd}}$ hexagonal number: $19$ (difference = 12)
- $4^{\text{th}}$ hexagonal number: $37$ (difference = 18)
- $5^{\text{th}}$ hexagonal number: $61$ (difference = 24)

The differences are $6, 12, 18, 24, \ldots$ (multiples of 6).

The next difference = $30$.

Next number: $61 + 30 = \mathbf{91}$

Formula: The $n$-th hexagonal number $= 1 + 6 + 12 + 18 + \cdots = 1 + 6\left(\dfrac{(n-1)n}{2}\right) = 3n^2 - 3n + 1$

For $n = 6$: $3(36) - 3(6) + 1 = 108 - 18 + 1 = 91$ ✓

Answer: They are called hexagonal numbers. The next number in the sequence is $\mathbf{91}$.

**Powers of 2: $1, 2, 4, 8, 16, 32, \ldots$

Pictorial way 1 — Squares/Rectangles:**
Start with 1 square. Each time, double the shape by placing an identical copy next to it.
- Step 0: 1 unit square
- Step 1: 2 unit squares (a $1 \times 2$ rectangle)
- Step 2: 4 unit squares (a $2 \times 2$ square)
- Step 3: 8 unit squares (a $2 \times 4$ rectangle)
- Step 4: 16 unit squares (a $4 \times 4$ square)

Each step doubles the number of squares.

Pictorial way 2 — Branching tree:
A tree where each branch splits into 2 at every level. At level $n$, there are $2^n$ branches.

**Powers of 3: $1, 3, 9, 27, 81, \ldots$

Pictorial way — Squares divided into 3 parts:
Start with 1 square. Divide it into 3 equal parts; each part is then divided into 3 more, and so on.
- Step 0: 1 square
- Step 1: 3 smaller squares
- Step 2: 9 smaller squares
- Step 3: 27 smaller squares

Pictorial way 2 — Branching tree:**
A tree where each branch splits into 3 at every level. At level $n$, there are $3^n$ branches.

Answer: Powers of 2 can be visualised by repeatedly doubling a rectangle or a branching tree with 2 branches at each node. Powers of 3 can be visualised by repeatedly tripling a square (like a Sierpinski carpet) or a branching tree with 3 branches at each node.

Given: The sequence $1,\ 1+2+1,\ 1+2+3+2+1,\ \ldots$

Computing the sums:
$$1 = 1 = 1^2$$
$$1+2+1 = 4 = 2^2$$
$$1+2+3+2+1 = 9 = 3^2$$
$$1+2+3+4+3+2+1 = 16 = 4^2$$

So the sums give square numbers.

Pictorial Explanation:

Consider an $n \times n$ square grid of dots. We can count the dots diagonally:
- The main diagonal (longest) has $n$ dots.
- The diagonals above and below it have $n-1, n-2, \ldots, 1$ dots each.

Counting all diagonals from top-left to bottom-right:
$$1 + 2 + 3 + \cdots + n + \cdots + 3 + 2 + 1 = n^2$$

This is because the diagonals of an $n \times n$ square grid, counted from one corner to the other, give exactly $1, 2, 3, \ldots, n, \ldots, 3, 2, 1$ dots, and together they fill the entire $n \times n$ square.

Answer: An $n \times n$ square grid, when its dots are counted along diagonals, gives the sum $1+2+\cdots+n+\cdots+2+1 = n^2$, confirming that these sums are perfect squares.

Given: The sum $S = 1 + 2 + 3 + \cdots + 99 + 100 + 99 + \cdots + 3 + 2 + 1$

Concept used: From the pictorial pattern established in Question 1:
$$1 + 2 + 3 + \cdots + n + \cdots + 3 + 2 + 1 = n^2$$

Working:
Here, the largest number in the sum is $100$, so $n = 100$.

$$S = 1 + 2 + 3 + \cdots + 100 + \cdots + 3 + 2 + 1 = 100^2$$

$$\boxed{S = 10000}$$

Pictorial reasoning: This sum counts all the dots in a $100 \times 100$ square grid when counted along diagonals, giving $100^2 = 10000$ dots total.

Answer: $1 + 2 + 3 + \cdots + 100 + \cdots + 3 + 2 + 1 = \mathbf{10000}$

All 1's sequence: $1, 1, 1, 1, 1, \ldots$

Part 1 — Adding the All 1's sequence up (cumulative sums):
$$1 = 1$$
$$1 + 1 = 2$$
$$1 + 1 + 1 = 3$$
$$1 + 1 + 1 + 1 = 4$$

We get the Counting Numbers (Natural Numbers): $1, 2, 3, 4, 5, \ldots$

Part 2 — Adding the All 1's sequence up and down:
$$1 = 1$$
$$1 + 1 + 1 = 3$$
$$1 + 1 + 1 + 1 + 1 = 5$$
$$1 + 1 + 1 + 1 + 1 + 1 + 1 = 7$$

We get the Odd Numbers: $1, 3, 5, 7, 9, \ldots$

Answer:
- Adding the All 1's sequence up gives the counting numbers: $1, 2, 3, 4, 5, \ldots$
- Adding the All 1's sequence up and down gives the odd numbers: $1, 3, 5, 7, 9, \ldots$

Counting numbers: $1, 2, 3, 4, 5, \ldots$

Cumulative sums:
$$1 = 1$$
$$1 + 2 = 3$$
$$1 + 2 + 3 = 6$$
$$1 + 2 + 3 + 4 = 10$$
$$1 + 2 + 3 + 4 + 5 = 15$$

We get the Triangular Numbers: $1, 3, 6, 10, 15, \ldots$

Pictorial Explanation:
When we add $1 + 2 + 3 + \cdots + n$, we are stacking rows of dots:
- Row 1: 1 dot
- Row 2: 2 dots
- Row 3: 3 dots
- $\vdots$
- Row $n$: $n$ dots

This forms a triangle shape, which is exactly why these are called triangular numbers.

The $n$-th triangular number $= \dfrac{n(n+1)}{2}$.

Answer: Adding the counting numbers up gives the triangular numbers $1, 3, 6, 10, 15, \ldots$ The pictorial explanation is that stacking rows of 1, 2, 3, ..., $n$ dots forms a triangle.

Given: Pairs of consecutive triangular numbers.

Computing:
$$1 + 3 = 4 = 2^2$$
$$3 + 6 = 9 = 3^2$$
$$6 + 10 = 16 = 4^2$$
$$10 + 15 = 25 = 5^2$$
$$15 + 21 = 36 = 6^2$$

We get the Square Numbers: $4, 9, 16, 25, 36, \ldots$ (i.e., $2^2, 3^2, 4^2, 5^2, \ldots$)

Why does this happen?

The $n$-th triangular number is $T_n = \dfrac{n(n+1)}{2}$.

$$T_{n-1} + T_n = \frac{(n-1)n}{2} + \frac{n(n+1)}{2} = \frac{n(n-1) + n(n+1)}{2} = \frac{n(2n)}{2} = n^2$$

Pictorial Explanation:
The $n$-th triangular number forms a right-angled triangle. If we take $T_{n-1}$ (a triangle with $n-1$ rows) and flip it upside down, it fits perfectly with $T_n$ (a triangle with $n$ rows) to form an $n \times n$ square.

Answer: Adding consecutive triangular numbers gives square numbers. This is because $T_{n-1} + T_n = n^2$, and pictorially, two consecutive triangles fit together to form a square.

Given: Cumulative sums of powers of 2.

Computing the sums:
$$1 = 1$$
$$1 + 2 = 3$$
$$1 + 2 + 4 = 7$$
$$1 + 2 + 4 + 8 = 15$$
$$1 + 2 + 4 + 8 + 16 = 31$$

The sequence is: $1, 3, 7, 15, 31, \ldots$

Adding 1 to each:
$$1 + 1 = 2$$
$$3 + 1 = 4$$
$$7 + 1 = 8$$
$$15 + 1 = 16$$
$$31 + 1 = 32$$

We get the Powers of 2: $2, 4, 8, 16, 32, \ldots$

Why does this happen?

Using the formula for the sum of a geometric series:
$$1 + 2 + 4 + \cdots + 2^{n-1} = 2^n - 1$$

So adding 1 gives $2^n - 1 + 1 = 2^n$, which is the next power of 2.

Pictorial reasoning: If you have a row of $2^n$ squares and fill them one by one doubling each time, you always need just one more square to complete the next power of 2.

Answer: The cumulative sums are $1, 3, 7, 15, 31, \ldots$ (one less than powers of 2). Adding 1 to each gives $2, 4, 8, 16, 32, \ldots$ — the powers of 2 — because $1 + 2 + 4 + \cdots + 2^{n-1} = 2^n - 1$.