Pascal’s Triangle – Binomial Theorem

We know the expansions of terms like (x + 2)²  and (x + 3)³. These are pretty simple to do, but sometimes we come across some expressions like (x+2)⁵. Expressions like these are hard to expand, we can use some tricks to simplify for example, 

(x+2)⁵ = (x+2)(x + 2)²(x+2)²

Researcher work on the binomial expansions which involve thousands of exponents. It becomes essential to know a simple way to expand them. This is still a lot of work, that’s where binomial theorem comes to our rescue. It allows us to expand any general expressions like (x+ a)ⁿ. Let’s look at this theorem in detail. 

Binomial Expression 

A binomial expression is defined as an expression that has two terms that are connected by operators like + or -.  For example, x + a, x – 6, and so on are examples of binomial expressions. Raising a binomial expression to a power greater than 3 is pretty hard and cumbersome. Let’s see some binomial expansions and try to find some pattern in them, 

(x + a)⁰ =                                                                                                              1

(x + a)¹ =                                                                                                           x + a 

(x + a)² =                                                                                                    x² + 2ax + a²

(x + a)³ =                                                                                             x³ + 3a²x + 3ax² + a³

Notice, we need to know the coefficients of these terms, and then it can be a little easier for us to describe their expansions. Pascals’ triangle can be used to generate these results really quickly. 

What is Pascal’s Triangle?

It is named after the famous Philosopher and Mathematician ‘Pascal’ who developed a pattern of numbers starting with 1 and the numbers beneath are the summation of the above numbers. To start making Pascal’s triangle, first, write down number 1. The second row is written down by two 1s again. Other rows are generated using the previous rows to make a triangle of numbers. Each row begins and ends with a 1. 

                                                                                            1
                                                                                      1           1
                                                                                1          x          1
                                                                          1        x         x          1

The numbers that are given by x are calculated by adding the numbers from the previous row, which lie on the left and right above the given position. The figure below demonstrates the process of building Pascal’s triangle. 

In this way, Pascal’s triangle can be generated. 

Question 1: Generate the sixth row of Pascal’s triangle. 

Answer: 

Question 2: Generate the tenth row of Pascal’s triangle. 

Answer: 

Pascal’s Triangles for expanding Binomial Expressions 

Let’s see the above equations again, 

(x + a)⁰ =                                                                                1

(x + a)¹ =                                                                             x + a 

(x + a)² =                                                                      x² + 2ax + a²

(x + a)³ =                                                               x³ + 3a²x + 3ax² + a³

We can conclude a few things from these equations for (x + a)ⁿ, 

1. There’s always one more term than the value of n.
2. For each term, the sum of exponents is always equal to n.
3. For the variable “x”, the exponent starts from n and keeps on decreasing till zero. Similarly, for “a” exponent starts from 0 and goes till n.
4. Coefficients of these terms are given by Pascal’s triangle.

Let’s see an example, suppose we want to expand (x + a)³ through this expansion concept. There should be four terms and the terms should have a decreasing exponent of “x” and an increasing exponent of “a” respectively. 

(x + a)³ = C₁x³ + C₂x²a + C₃xa² + C₄a³

The values C_1, C₂, C₃, and C₄ are coefficients, we will figure out the coefficients with Pascal’s triangle. Let’s see Pascal’s triangle with n + 1 rows, 

The values in the last row give us the value of coefficients C₁, C₂, C₃, and C₄. 

Here, C₁ = 1, C₂ = 3, C₃ = 3 and C₄ = 1. 

So, in this way, we can expand our binomial expressions. 

For any binomial expression, (x + a)ⁿ the expansions is given by, 

(a + b)ⁿ = c₀aⁿb⁰ + c₁a^n-1b¹ + c₂a^n-2b² + …. + c_n-1a¹b^n-1 + c_na⁰bⁿ

The coefficients are given by the n+1 row of the Pascal’s triangle. 

Sample Problems

Question 1: Expand and verify (a + b)². 

Solution:

First write the generic expressions without the coefficients.  

 (a + b)² = c₀a²b⁰ + c₁a¹b¹ + c₂a⁰b²

Now let’s build a Pascal’s triangle for 3 rows to find out the coefficients.  

The values of the last row give us the value of coefficients. 

c₀ = 1, c₁ = 2, c₂ =1

 (a + b)² = a²b⁰ + 2a¹b¹ + a⁰b²

Thus verified. 

Question 2: Expand (a + b)⁴. 

Solution:

First write the generic expressions without the coefficients.  

 (a + b)⁴ = c₀a⁴b⁰ + c₁a³b¹ + c₂a²b² + c₃a¹b³ + c₄a⁰b⁴

Now let’s build a Pascal’s triangle for 5 rows to find out the coefficients.  

The values of the last row give us the value of coefficients. 

c₀ = 1, c₁ = 4, c₂ = 6, c₃ = 4 and c₄ =1. 

Thus, (a + b)⁴ = a⁴b⁰ + 4a³b¹ +6a²b² + 4a¹b³ + a⁰b⁴

Question 3: Expand (a + b)⁵. 

Solution:

First write the generic expressions without the coefficients.  

 (a + b)⁵ = c₀a⁵b⁰ + c₁a⁴b¹ + c₂a³b² + c₃a²b³ + c₄a¹b⁴ + c₅a⁰b⁵

Now let’s build a Pascal’s triangle for 6 rows to find out the coefficients.  

The values of the last row give us the value of coefficients. 

c₀ = 1, c₁ = 5, c₂ = 10, c₃ = 10, c₄ =5 and c₅ = 1. 

(a + b)⁵ = a⁵b⁰ + 5a⁴b¹ + 10a³b² + 10a²b³ + 5a¹b⁴ + a⁰b⁵

Question 4: Expand (a + b)⁶. 

Solution:

First write the generic expressions without the coefficients.  

 (a + b)⁶ = c₀a⁶b⁰ + c₁a⁵b¹ + c₂a⁴b² + c₃a³b³ + c₄a²b⁴ + c₅a¹b⁵ + c₆a⁰b⁶ 

Now let’s build a Pascal’s triangle for 7 rows to find out the coefficients.  

The values of the last row give us the value of coefficients. 

c₀ = 1, c₁ = 6, c₂ = 15, c₃ = 20, c₄ =15, c₅ = 6 and c₆ = 1. 

(a + b)⁶ = 1a⁶b⁰ + 6a⁵b¹ + 15a⁴b² + 20a³b³ + 15a²b⁴ + 6a¹b⁵ + 1a⁰b⁶