Algebra As Patterns

If you think of Mathematics as a language, then Algebra would be that part of the language which describes various patterns around us. If there is any repetitive pattern, then we can use algebra to simplify this and have a general expression to describe this pattern.

ALGEBRA PATTERNS

To understand this connection between patterns and algebra, let’s try something. We can use pencils to construct a simple pattern and understand how to create a general expression to describe the entire pattern. You need a lot of pencils for this. It would help if they are of similar height.

Find a solid surface and arrange two pencils parallel to each other with some space in between them. Add a second layer on top of it and another on top of that as shown in the image given below.PatternThere is a total of six pencils in this arrangement. Here, there are three layers and each layer has a fixed count of two pencils. The number of pencils in each layer never varies, but the number of layers you wish to build is entirely up to you.

Current Number of Layers = 4

Number of Pencils per Layer = 2

Total Number of Pencils = 2 x 4 = 8

What if you increase the number of layers to 10? What if you keep building up to a layer of 100? Can you literally sit and stack those many layers? Obviously not. Instead, let’s try to calculate.

Number of Layers = 100

Number of Pencils per Layer = 2

Total Number of Pencils = 2 x 100 = 200

There is an obvious pattern here. A single level has 2 pencils and this is always constant, regardless of the number of levels built. So to get the total number of pencils, we just have to multiply 2 (the number of pencils per level) with the number of levels built. For example, to construct 30 levels, you will need 2 multiplied 30 times which is 60 pencils.

To make a building of ‘x’ number of levels, according to the previous calculation, we will require 2 multiplied ‘x’ times which is 2x number of pencils. We just created an algebraic expression based on patterns.

Let’s look at another pattern. Say we have a triangle.PatternFlip the triangle downwards and complete this image to form a complete triangle as given below.PatternThe pattern is still a triangle but the number of smaller triangles increases to 4. This bigger triangle now has two rows. What happens when we increase the number of rows, and then fill in the gaps, with smaller triangles, to complete that big triangle?

As we proceed to the 3rd row, how many smaller triangles do we have in total now? Now, this triangle has 9 smaller triangles.PatternWhat is the pattern here?

When,

r = 1, total no of triangles, ‘n’ = 1

When,

r = 2?  n = 4

And when,

r = 3, n = 9

You can see that as the size of the triangle increases the number of smaller triangles also increases. This means that in this case n is just equal to r² which is 1², 2², 3² = 1, 4, 9…

So how do we represent this triangular pattern algebraically? r²! That’s it! We have just represented patterns algebraically.

The sequence of odd numbers can be written as 1, 3, 5, 7, 9……………………….. (2n – 1). Now if we substitute the values of ‘n’ in the expression beginning from one for odd numbers i.e. 2n – 1, the nth odd number can be easily calculated. To find the 11th odd number, substitute n= 11, the result is 21 i.e. 11th odd number. Similarly, the 100th odd number is 199. Thus, the algebraic expression representing a set of odd numbers is 2n -1. It must be noted that n is a set of whole numbers in this case.

Similar to patterns in numbers we can figure out the pattern in figures. For example, consider the following figures:

Algebra - Patterns And Algebra

In the figure given above, the first image is a pentagon with five sides. In the next figure, two pentagons are joined end to end and the total number of sides is 9, next figure the total number of sides is 13 and so on. It is observed that every other figure is having 4 sides extra as compared to the previous one. Thus, the algebraic pattern that would define this sequence exactly is ‘4n + 1’, where n is any natural number. Thus if we substitute n = 3, we get the number of sides equal to 13. The 10th pattern in this sequence will have sides equal to 4 × 10 + 1 i.e.  41 sides.

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Practise This Question

Observe the pattern given below

 

 

Then term at nth position will be given by expression