Patterns in Mathematics Class 6 Solutions

Many answers are possible. Good examples include:

• comparing prices and checking change while shopping;
• measuring ingredients and cooking time;
• reading a clock and planning how long a journey will take;
• measuring length, area and material while making furniture or a room;
• using scores, averages and distances in sports;
• reading temperature and rainfall information.

This is open-ended. An answer is valid when it names a real situation and explains what is counted, measured or compared.

Mathematics lets people describe patterns precisely and predict what may happen. Engineers use calculation and geometry to design safe bridges and buildings. Scientists use measurement and statistics to test ideas. Computers and mobile phones depend on mathematical logic. Calendars and clocks use repeating astronomical patterns, while transport systems use distance, speed and timing.

A strong response gives two or three examples and connects each invention or system to a mathematical idea.

For triangular numbers, the next additions are 8, 9 and 10. For example, 28 + 8 = 36 and 36 + 9 = 45.

The next picture must use the next number of objects in that sequence:

• All 1s: one object again.
• Counting: 6 objects.
• Odd and even: 11 objects and 12 objects.
• Triangular: 21 dots in 6 rows.
• Squares: 36 dots in a 6 × 6 grid.
• Cubes: 216 small cubes in a 6 × 6 × 6 cube.

The exact drawing style may differ; the arrangement and count must follow the same rule as the textbook figures.

Triangular numbers can be arranged in rows that form a triangle: 1; 1 + 2 = 3; 1 + 2 + 3 = 6, and so on. Square numbers form equal rows and columns, such as 4 = 2 × 2 and 9 = 3 × 3. Cube numbers form equal length, width and height, such as 8 = 2 × 2 × 2 and 27 = 3 × 3 × 3.

For the triangle, arrange dots in rows of 1, 2, 3, 4, 5, 6, 7 and 8. Their total is 36. For the square, arrange the same 36 dots in 6 equal rows of 6.

The increases are 6, 12 and 18. The amount added grows by 6 each time, so add 24 next.

37 + 24 = 61

The next hexagonal number is 61.

For powers of 2, start with one square and replace every square with two squares at each stage: 1, 2, 4, 8, 16. For powers of 3, replace each object with three objects: 1, 3, 9, 27. A branching tree is another correct picture: every branch splits into two or three new branches.

This is a creative question. Any drawing that makes repeated doubling or tripling visible is valid.

The sum of the first n odd numbers is an n by n square. With 10 odd numbers, the square has side 10.

1 + 3 + 5 + ... + 19 = 10 × 10 = 100

Using the same square pattern, 100 odd-number layers make a 100 by 100 square.

1 + 3 + 5 + ... + 199 = 100 × 100 = 10,000

Take the fourth expression: 1 + 2 + 3 + 4 + 3 + 2 + 1. Arrange these as seven rows, centred from the shortest to the longest and back. Move the extra parts of the shorter rows into the missing spaces. They fill a 4 × 4 square exactly, so the sum is 16.

The same rearrangement works for every size: a peak of n gives an n by n square.

An up-and-down sum peaking at 100 forms a 100 × 100 square.

1 + 2 + ... + 99 + 100 + 99 + ... + 2 + 1 = 100² = 10,000

Partial sums of all 1s are 1, 1 + 1, 1 + 1 + 1, ... so they give 1, 2, 3, 4, ..., the counting numbers.

Going up and down uses 1 object, then 3 objects, then 5 objects, and so on. It gives 1, 3, 5, 7, ..., the odd numbers.

They produce triangular numbers:

1; 1 + 2 = 3; 1 + 2 + 3 = 6; 1 + 2 + 3 + 4 = 10

A picture uses rows of 1, 2, 3, 4, ... dots. Each new row adds the next counting number and keeps the triangular shape.

The sums are:

1 + 3 = 4,   3 + 6 = 9,   6 + 10 = 16,   10 + 15 = 25

These are square numbers. Two consecutive triangular dot arrangements fit together to make a square: T_n + T_n+1 = (n + 1)².

The partial sums are 1, 3, 7, 15, 31, .... Adding 1 gives 2, 4, 8, 16, 32, ..., which are powers of 2.

Each partial sum is one less than the next power of 2. For example:

1 + 2 + 4 + 8 = 15 = 16 − 1

Using 1, 3, 6, 10, 15, ...:

6×1+1=7, 6×3+1=19, 6×6+1=37, 6×10+1=61, 6×15+1=91

This continues the centred hexagonal sequence after its first term 1. Each new triangular stage accounts for six equal triangular sectors around a centre dot.

The partial sums are:

1; 1+7=8; 1+7+19=27; 1+7+19+37=64

So the sequence is 1, 8, 27, 64, ..., the cube numbers. Geometrically, each new centred-hexagonal layer is the extra set of unit cubes needed to grow an n × n × n cube into the next cube.

One relationship is that the difference between consecutive square numbers is always the next odd number:

4−1=3,   9−4=5,   16−9=7,   25−16=9

In a dot picture, an n × n square becomes an (n+1) × (n+1) square by adding one row and one column. After avoiding the double-counted corner, that border contains 2n+1 dots, an odd number.

Other correctly explained relationships are also valid.

• Regular polygons: add one side and one corner each time.
• Complete graphs: add one point and join it to every existing point.
• Stacked squares: increase the stack so the total small-square count follows 1, 4, 9, 16, 25.
• Stacked triangles: add a new layer so the small-triangle count follows 1, 4, 9, 16, 25.
• Koch snowflake: replace every line segment with four shorter segments forming a bump.

• After the decagon comes a regular 11-sided polygon.
• After K₆ comes K₇: seven points with every pair joined.
• The next stacked square stage contains 36 little squares.
• The next stacked triangle stage contains 36 little triangles.
• The next Koch stage replaces each of the current segments with the same four-segment bump.

The first four can be sketched directly. The Koch figure is possible to draw, but it requires many small accurate segments, so a ruler or drawing software helps.

Both counts are 3, 4, 5, 6, 7, 8, 9, 10, .... Every side meets the next side at one corner, so a closed polygon has exactly as many corners as sides.

K₂, K₃, K₄, K₅ and K₆ have 1, 3, 6, 10 and 15 lines. These are triangular numbers.

Each new point connects to all earlier points. The added line counts are therefore 2, then 3, then 4, then 5, producing the successive triangular totals.

The totals are 1, 4, 9, 16, 25, .... They are square numbers because the nth arrangement can be rearranged into n rows of n little squares, giving n × n.

The totals are also 1, 4, 9, 16, 25, .... The row counts go up and down:

1;   1+2+1;   1+2+3+2+1;   ...

As shown earlier, a counting-up-and-down sum peaking at n equals n².

The counts are 3, 12, 48, 192, 768, .... Every old segment is replaced by four new segments, so the count is multiplied by 4 at each stage.

3,   3×4,   3×4²,   3×4³,   3×4⁴, ...