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Patterns in Mathematics - What is Mathematics?

Grade 6CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A pattern is a recurring sequence or arrangement that follows a specific rule. In Grade 6, we study patterns in numbers, shapes, and dot arrangements.

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Number Patterns: Sequences where numbers increase or decrease by a fixed value or follow a specific logic. For example, in the sequence 3,6,9,12,…3, 6, 9, 12, \dots, each term is a multiple of 33.

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Square Numbers: These are numbers that can be arranged in a square pattern of dots. A square number is obtained by multiplying a number by itself (n×nn \times n). Examples: 1,4,9,16,25,…1, 4, 9, 16, 25, \dots.

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Triangular Numbers: These are numbers that can be arranged as dots forming an equilateral triangle. The sequence is 1,3,6,10,15,…1, 3, 6, 10, 15, \dots. The nthn^{th} triangular number is the sum of the first nn natural numbers.

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Odd Number Summation: A fascinating pattern where the sum of the first nn odd numbers is always equal to n2n^2. For example, 1+3+5=91 + 3 + 5 = 9, which is 323^2.

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Generalizing Rules: We use variables (like nn) to represent the position of a term. If a pattern increases by 44 each time starting from 44, the nthn^{th} term is 4n4n.

📐Formulae

nth Square Number=n×n=n2n^{th} \text{ Square Number} = n \times n = n^2

nth Triangular Number=n(n+1)2n^{th} \text{ Triangular Number} = \frac{n(n + 1)}{2}

Sum of first n odd numbers=1+3+5+⋯+(2n−1)=n2\text{Sum of first } n \text{ odd numbers} = 1 + 3 + 5 + \dots + (2n-1) = n^2

General term of an Arithmetic Pattern with difference d and first term a=a+(n−1)d\text{General term of an Arithmetic Pattern with difference } d \text{ and first term } a = a + (n-1)d

💡Examples

Problem 1:

Find the 12th12^{th} term in the number pattern: 5,10,15,20,…5, 10, 15, 20, \dots

Solution:

12×5=6012 \times 5 = 60

Explanation:

The pattern follows the rule of multiples of 55. The nthn^{th} term is 5n5n. For n=12n=12, the term is 5×12=605 \times 12 = 60.

Problem 2:

Calculate the 5th5^{th} triangular number and represent it using the formula.

Solution:

5(5+1)2=5×62=15\frac{5(5 + 1)}{2} = \frac{5 \times 6}{2} = 15

Explanation:

Using the formula for the nthn^{th} triangular number where n=5n=5, we calculate the sum of the first 5 natural numbers (1+2+3+4+51+2+3+4+5), which equals 1515.

Problem 3:

Without actual addition, find the sum of the first 88 odd numbers: 1+3+5+7+9+11+13+151 + 3 + 5 + 7 + 9 + 11 + 13 + 15.

Solution:

82=648^2 = 64

Explanation:

The sum of the first nn odd numbers is given by the pattern n2n^2. Since there are 88 terms, the sum is 8×8=648 \times 8 = 64.

Problem 4:

In a matchstick pattern, 11 square requires 44 sticks, 22 squares require 77 sticks, and 33 squares require 1010 sticks. Find the rule for nn squares.

Solution:

Rule=3n+1\text{Rule} = 3n + 1

Explanation:

For 11 square: 3(1)+1=43(1) + 1 = 4. For 22 squares: 3(2)+1=73(2) + 1 = 7. For 33 squares: 3(3)+1=103(3) + 1 = 10. Thus, the number of sticks for nn squares is 3n+13n + 1.