Patterns in Mathematics — NCERT Solutions

Given: Triangular numbers: $1, 3, 6, 10, 15, 21, \ldots$

Computing $6 \times T_n + 1$:
$$6 \times 1 + 1 = 7$$
$$6 \times 3 + 1 = 19$$
$$6 \times 6 + 1 = 37$$
$$6 \times 10 + 1 = 61$$
$$6 \times 15 + 1 = 91$$
$$6 \times 21 + 1 = 127$$

The sequence is: $7, 19, 37, 61, 91, 127, \ldots$

This is the sequence of Hexagonal Numbers (starting from the 2nd term): $1, 7, 19, 37, 61, 91, \ldots$

Why does this happen?

The $n$-th hexagonal number is $H_n = 3n^2 - 3n + 1$.

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

$$6 \times T_{n-1} + 1 = 6 \times \frac{(n-1)n}{2} + 1 = 3n(n-1) + 1 = 3n^2 - 3n + 1 = H_n$$

Pictorial Explanation:
A hexagonal arrangement can be seen as 6 triangular arrangements placed around a central dot. So 6 triangles + 1 centre = hexagon.

Answer: Multiplying triangular numbers by 6 and adding 1 gives the hexagonal numbers $7, 19, 37, 61, \ldots$ This is because a hexagon can be formed by placing 6 triangular arrangements around a central point.

Given: Hexagonal numbers: $1, 7, 19, 37, 61, \ldots$

Computing cumulative sums:
$$1 = 1 = 1^3$$
$$1 + 7 = 8 = 2^3$$
$$1 + 7 + 19 = 27 = 3^3$$
$$1 + 7 + 19 + 37 = 64 = 4^3$$
$$1 + 7 + 19 + 37 + 61 = 125 = 5^3$$

We get the Cube Numbers: $1, 8, 27, 64, 125, \ldots$

Why does this happen?

The $n$-th hexagonal number is $H_n = 3n^2 - 3n + 1$.

$$\sum_{k=1}^{n} H_k = \sum_{k=1}^{n} (3k^2 - 3k + 1) = 3\cdot\frac{n(n+1)(2n+1)}{6} - 3\cdot\frac{n(n+1)}{2} + n$$
$$= \frac{n(n+1)(2n+1)}{2} - \frac{3n(n+1)}{2} + n = \frac{n(n+1)(2n+1) - 3n(n+1) + 2n}{2}$$
$$= \frac{n[(n+1)(2n+1) - 3(n+1) + 2]}{2} = \frac{n[2n^2 + 3n + 1 - 3n - 3 + 2]}{2} = \frac{n(2n^2)}{2} = n^3$$

Pictorial Explanation using a cube:
An $n \times n \times n$ cube can be built layer by layer. The $k$-th layer (a square slab) contributes $H_k = 3k^2 - 3k + 1$ unit cubes (the $k$-th hexagonal number). Adding all layers from 1 to $n$ gives the total $n^3$ unit cubes.

Answer: Adding up hexagonal numbers gives cube numbers $1, 8, 27, 64, 125, \ldots$ because each hexagonal number represents one layer of a cube, and stacking $n$ such layers builds an $n \times n \times n$ cube.

Given: Various number sequences in Table 1.

Some interesting patterns and relations:

Pattern 1: Every even-positioned triangular number is also divisible by 3.

• $T_2 = 3,\ T_4 = 10,\ T_6 = 21,\ T_8 = 36$ — $T_2, T_6$ are divisible by 3.

Pattern 2: The sum of the first $n$ counting numbers equals the $n$-th triangular number:
$$1 + 2 + 3 + \cdots + n = T_n = \frac{n(n+1)}{2}$$

Pattern 3: The difference between consecutive square numbers gives odd numbers:
$$4 - 1 = 3,\ 9 - 4 = 5,\ 16 - 9 = 7,\ 25 - 16 = 9, \ldots$$
This is because $(n+1)^2 - n^2 = 2n + 1$, which is always odd.

Pattern 4: Every square number is the sum of two consecutive triangular numbers:
$$n^2 = T_{n-1} + T_n$$
(As shown in Question 5 above.)

Pattern 5: The Virahānka (Fibonacci) numbers appear in Pascal's triangle along the diagonals.

Pattern 6: Powers of 2 are always even except for $2^0 = 1$.

Answer: There are many beautiful patterns among the sequences. Three key ones are: (i) differences of consecutive squares are odd numbers, (ii) sums of consecutive triangular numbers are squares, and (iii) cumulative sums of odd numbers are squares. Each can be explained pictorially using dot arrangements or grid diagrams.

Given: Shape sequences in Table 3.

Patterns in each sequence:

1. Regular Polygons: Each shape has one more side than the previous. The sequence goes: equilateral triangle (3 sides), square (4 sides), regular pentagon (5 sides), regular hexagon (6 sides), and so on. The number of sides increases by 1 each time.

2. Complete Graphs: Each new shape adds one more point (node), and the new point is connected to all existing points. The number of lines increases by 1, 2, 3, 4, ... at each step.

3. Stacked Squares: Each shape is formed by adding one more row and one more column of unit squares. The total number of unit squares follows the sequence of square numbers: $1, 4, 9, 16, \ldots$

4. Stacked Triangles: Each shape is formed by adding one more row of small triangles. The total number of small triangles follows the sequence of square numbers: $1, 4, 9, 16, \ldots$

5. Koch Snowflake: Each shape is formed by replacing every line segment with a 'speed bump' shape (a line with a triangular bump in the middle). The number of line segments multiplies by 4 at each step.

Answer: Each sequence in Table 3 follows a clear geometric rule — adding sides, adding nodes and connections, adding rows of squares or triangles, or replacing segments with bumps.

Working:

1. Regular Polygons:

• Rule: Each shape has one more side than the previous, with all sides equal and all angles equal.
• Next shape after regular hexagon (6 sides): Regular Heptagon (7 equal sides).
• ✓ Can be drawn.

2. Complete Graphs:

• Rule: Add one new point and connect it to all existing points with straight lines.
• If the previous shape had 4 points (6 lines), the next has 5 points connected to each other = $\binom{5}{2} = 10$ lines.
• ✓ Can be drawn.

3. Stacked Squares:

• Rule: Each shape is an $n \times n$ grid of unit squares.
• Next shape after $4 \times 4$ grid: $5 \times 5$ grid (25 unit squares).
• ✓ Can be drawn.

4. Stacked Triangles:

• Rule: Each shape is formed by stacking rows of small equilateral triangles, with each row having more triangles.
• Next shape: Add one more row.
• ✓ Can be drawn.

5. Koch Snowflake:

• Rule: Replace each line segment '—' with a 'speed bump' $\wedge$ shape (four segments instead of one).
• Next shape: Apply the replacement to every segment of the current shape.
• ✓ Can be drawn (though it becomes very detailed).

Answer: All next shapes can be drawn by following the respective rules. The rules are: add one side (polygons), add one connected node (complete graphs), add one row/column (stacked squares/triangles), and replace each segment with a bump (Koch snowflake).

Given: Sequence of Regular Polygons: triangle, square, pentagon, hexagon, heptagon, ...

Number of sides:
$$3, 4, 5, 6, 7, 8, \ldots$$
This is the counting numbers starting from 3.

Number of corners (vertices):
$$3, 4, 5, 6, 7, 8, \ldots$$
This is also the counting numbers starting from 3 — the same sequence!

Why are they the same?

In any polygon, the number of corners (vertices) is always equal to the number of sides. This is because each side connects two consecutive corners, and as you go around the polygon, each corner is the meeting point of exactly two sides. So a polygon with $n$ sides always has exactly $n$ corners.

Answer: Both the number of sides and the number of corners give the sequence $3, 4, 5, 6, 7, \ldots$ (counting numbers starting at 3). They are equal because in any polygon, the number of sides equals the number of vertices.

Given: Sequence of Complete Graphs with $1, 2, 3, 4, 5, \ldots$ points.

Counting lines (edges):

• 1 point: $0$ lines
• 2 points: $1$ line
• 3 points: $3$ lines
• 4 points: $6$ lines
• 5 points: $10$ lines
• 6 points: $15$ lines

The sequence is: $0, 1, 3, 6, 10, 15, \ldots$

This is the Triangular Numbers (starting from 0): $T_0, T_1, T_2, T_3, \ldots$

Why does this happen?

In a complete graph with $n$ points, every point is connected to every other point. The number of lines is the number of ways to choose 2 points from $n$ points:
$$\text{Number of lines} = \binom{n}{2} = \frac{n(n-1)}{2}$$

This is the $(n-1)$-th triangular number:
$$\frac{n(n-1)}{2} = T_{n-1}$$

When we add the $n$-th point, it connects to all $(n-1)$ existing points, adding $(n-1)$ new lines. So the number of lines increases by $1, 2, 3, 4, \ldots$ — exactly the counting numbers — giving triangular numbers.

Answer: The number of lines in complete graphs gives the triangular numbers $0, 1, 3, 6, 10, 15, \ldots$ because each new point connects to all previous points, adding one more line each time.

Given: Sequence of Stacked Squares.

Counting unit squares:

• Shape 1: $1 \times 1 = 1$ square
• Shape 2: $2 \times 2 = 4$ squares
• Shape 3: $3 \times 3 = 9$ squares
• Shape 4: $4 \times 4 = 16$ squares
• Shape 5: $5 \times 5 = 25$ squares

The sequence is: $1, 4, 9, 16, 25, \ldots$

This is the Square Numbers sequence.

Why does this happen?

The $n$-th shape in the sequence is an $n \times n$ grid. It contains $n$ rows, each with $n$ unit squares, giving a total of $n \times n = n^2$ unit squares. Since $n^2$ is the $n$-th square number, the sequence of counts gives the square numbers.

Answer: The number of unit squares in each stacked square shape gives the square numbers $1, 4, 9, 16, 25, \ldots$ because the $n$-th shape is an $n \times n$ grid containing $n^2$ unit squares.

Given: Sequence of Stacked Triangles.

Counting unit triangles (using the hint):

In the $n$-th shape:

• Row 1 (top): 1 triangle
• Row 2: 3 triangles
• Row 3: 5 triangles
• Row $k$: $(2k - 1)$ triangles

Total triangles in the $n$-th shape:
$$1 + 3 + 5 + \cdots + (2n-1) = n^2$$

Verification:

• Shape 1: $1 = 1^2 = 1$ ✓
• Shape 2: $1 + 3 = 4 = 2^2$ ✓
• Shape 3: $1 + 3 + 5 = 9 = 3^2$ ✓
• Shape 4: $1 + 3 + 5 + 7 = 16 = 4^2$ ✓

The sequence is: $1, 4, 9, 16, 25, \ldots$

This is the Square Numbers sequence.

Why does this happen?

Each row $k$ of the stacked triangle contains $(2k-1)$ small triangles (an odd number). The sum of the first $n$ odd numbers equals $n^2$ (as established in the pattern of adding odd numbers giving square numbers).

Answer: The number of unit triangles in each stacked triangle shape gives the square numbers $1, 4, 9, 16, \ldots$ because row $k$ has $(2k-1)$ triangles, and the sum of the first $n$ odd numbers is $n^2$.

Given: Koch Snowflake sequence. Each line segment is replaced by 4 segments (a 'speed bump' replaces '—' with a shape made of 4 segments).

Counting line segments:

• Shape 1 (equilateral triangle): 3 line segments
• Shape 2: Each of the 3 segments is replaced by 4 segments: $3 \times 4 = 12$ segments
• Shape 3: Each of the 12 segments is replaced by 4 segments: $12 \times 4 = 48$ segments
• Shape 4: $48 \times 4 = 192$ segments
• Shape 5: $192 \times 4 = 768$ segments

The sequence: $3, 12, 48, 192, 768, \ldots$

Pattern: Each term is 4 times the previous term.
$$3 = 3 \times 4^0,\quad 12 = 3 \times 4^1,\quad 48 = 3 \times 4^2,\quad 192 = 3 \times 4^3, \ldots$$

The $n$-th term $= 3 \times 4^{n-1}$.

This is 3 times the Powers of 4: $3 \times 1,\ 3 \times 4,\ 3 \times 16,\ 3 \times 64, \ldots$

Answer: The number of line segments in the Koch Snowflake shapes gives the sequence $3, 12, 48, 192, \ldots$ which is 3 times the powers of 4 (i.e., $3 \times 4^{n-1}$ for the $n$-th shape), because each step multiplies the number of segments by 4.