# Nakafa Learning Content
> For AI agents: use [llms.txt](https://nakafa.com/llms.txt) for the site index. Markdown versions are available by appending `.md` to content URLs or sending `Accept: text/markdown`.
URL: https://nakafa.com/en/subjects/mathematics/combinatorics/binomial-newton
Source: https://raw.githubusercontent.com/nakafaai/nakafa.com/refs/heads/main/packages/contents/material/lesson/mathematics/combinatorics/binomial-newton/en.mdx
Learn binomial theorem to expand (x+y)^n quickly. Learn coefficients, constant terms, and problem-solving with worked examples and practice problems.
---
## What is Binomial Newton?
Have you ever wondered how to quickly calculate the result of $$(x + y)^{10}$$ without having to multiply over and over again? **Binomial Newton** is a mathematical technique that allows us to expand the form $$(x + y)^n$$ into a sum of simpler terms.
Visible text: Have you ever wondered how to quickly calculate the result of without having to multiply over and over again? **Binomial Newton** is a mathematical technique that allows us to expand the form into a sum of simpler terms.
Imagine it like unwrapping a layered gift. Each layer we open reveals a certain pattern that is consistent and predictable. Similarly with binomial Newton, each power has a unique coefficient pattern that can be calculated with the same formula.
Let's look at the basic patterns for the first few powers:
Component: MathContainer
Children:
```math
(x + y)^0 = 1
```
```math
(x + y)^1 = x + y
```
```math
(x + y)^2 = x^2 + 2xy + y^2
```
```math
(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3
```
```math
(x + y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4
```
From this pattern, we can see that each term has **certain coefficients** that follow clear mathematical rules.
## General Formula and Binomial Coefficients
The general formula for binomial Newton can be written as:
```math
(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k
```
Where $$\binom{n}{k}$$ is the **binomial coefficient** calculated with the formula:
Visible text: Where is the **binomial coefficient** calculated with the formula:
```math
\binom{n}{k} = \frac{n!}{k!(n-k)!}
```
This binomial coefficient is also known as "n choose k" because it shows how many ways to choose k objects from n available objects.
The complete expansion form can be written as:
```math
(x + y)^n = \binom{n}{0}x^n + \binom{n}{1}x^{n-1}y + \binom{n}{2}x^{n-2}y^2 + \cdots + \binom{n}{n}y^n
```
Each term in the expansion has the structure $$\binom{n}{k}x^{n-k}y^k$$ where the powers of $$x$$ and $$y$$ always sum to $$n$$.
Visible text: Each term in the expansion has the structure where the powers of and always sum to .
## Finding Specific Coefficients
One important application of binomial Newton is finding coefficients of specific terms without having to expand the entire expression.
Suppose we want to find the coefficient of $$x^2$$ in the expansion of $$(1 - x)^{2014}$$.
Visible text: Suppose we want to find the coefficient of in the expansion of .
First, we rewrite it in standard binomial form with $$a = 1$$, $$b = -x$$, and $$n = 2014$$:
Visible text: First, we rewrite it in standard binomial form with , , and :
```math
(1 + (-x))^{2014} = \sum_{k=0}^{2014} \binom{2014}{k} (1)^{2014-k} (-x)^k
```
To get the term containing $$x^2$$, we need $$k = 2$$:
Visible text: To get the term containing , we need :
```math
\binom{2014}{2} (1)^{2012} (-x)^2 = \binom{2014}{2} \cdot 1 \cdot x^2 = \binom{2014}{2} x^2
```
Calculating the binomial coefficient:
Component: MathContainer
Children:
```math
\binom{2014}{2} = \frac{2014!}{2!(2014-2)!} = \frac{2014 \times 2013}{2 \times 1}
```
```math
= \frac{4{,}053{,}182}{2} = 2{,}026{,}591
```
Therefore, the coefficient of $$x^2$$ is $$2{,}026{,}591$$.
Visible text: Therefore, the coefficient of is .
## Finding the Constant Term
The constant term is a term that does not contain any variables. To find it, we need to identify the term where the power of all variables equals zero.
**Example:** Determine the constant term of $$\left(3x^3 - \frac{2}{x}\right)^8$$.
Visible text: **Example:** Determine the constant term of .
We write it in binomial form with $$a = 3x^3$$ and $$b = -\frac{2}{x}$$:
Visible text: We write it in binomial form with and :
```math
\left(3x^3 - \frac{2}{x}\right)^8 = \sum_{k=0}^{8} \binom{8}{k} (3x^3)^{8-k} \left(-\frac{2}{x}\right)^k
```
The general term is:
Component: MathContainer
Children:
```math
\binom{8}{k} (3x^3)^{8-k} \left(-\frac{2}{x}\right)^k = \binom{8}{k} \cdot (3)^{8-k} \cdot (x^3)^{8-k} \cdot (-2)^k \cdot (x^{-1})^k
```
```math
= \binom{8}{k} \cdot 3^{8-k} \cdot (-2)^k \cdot x^{3(8-k)} \cdot x^{-k}
```
```math
= \binom{8}{k} \cdot 3^{8-k} \cdot (-2)^k \cdot x^{24-3k-k} = \binom{8}{k} \cdot 3^{8-k} \cdot (-2)^k \cdot x^{24-4k}
```
For the constant term, the power of $$x$$ must be zero:
Visible text: For the constant term, the power of must be zero:
```math
24 - 4k = 0 \Rightarrow 4k = 24 \Rightarrow k = 6
```
Substituting $$k = 6$$:
Visible text: Substituting :
Component: MathContainer
Children:
```math
\binom{8}{6} \cdot 3^{8-6} \cdot (-2)^6 \cdot x^0 = \binom{8}{6} \cdot 3^2 \cdot (-2)^6
```
```math
= \frac{8!}{6! \cdot 2!} \cdot 9 \cdot 64 = \frac{8 \times 7}{2 \times 1} \cdot 9 \cdot 64
```
```math
= 28 \cdot 9 \cdot 64 = 252 \cdot 64 = 16{,}128
```
Note that $$(-2)^6 = 64$$ because even powers always produce positive values, just like $$(+2)^6 = 64$$.
Visible text: Note that because even powers always produce positive values, just like .
Therefore, the constant term is $$16{,}128$$.
Visible text: Therefore, the constant term is .
## Problem Solving Strategy
When facing binomial Newton problems, follow these systematic steps:
1. **Identify the components** in the form $$(a + b)^n$$ and clearly determine the values of a, b, and n.
2. **Determine the type of term** being sought, whether it's a specific coefficient, constant term, or term with a specific power.
3. **Use the general term formula** $$\binom{n}{k} a^{n-k} b^k$$ and adjust according to the required conditions.
4. **Calculate carefully** the binomial coefficient values and other arithmetic operations.
Visible text: 1. **Identify the components** in the form and clearly determine the values of a, b, and n.
2. **Determine the type of term** being sought, whether it's a specific coefficient, constant term, or term with a specific power.
3. **Use the general term formula** and adjust according to the required conditions.
4. **Calculate carefully** the binomial coefficient values and other arithmetic operations.
**Example Strategy Application:**
Determine the coefficient of $$x^5$$ in the expansion of $$(2x + 3)^8$$.
Visible text: Determine the coefficient of in the expansion of .
1. **Identify components**
From $$(2x + 3)^8$$, we get:
- $$a = 2x$$
- $$b = 3$$
- $$n = 8$$
2. **Determine the type of term**
We are looking for the coefficient of the term containing $$x^5$$.
3. **Use the general term formula**
General term: $$\binom{8}{k} (2x)^{8-k} (3)^k$$
Expanding the general term:
```math
\binom{8}{k} (2x)^{8-k} (3)^k = \binom{8}{k} \cdot 2^{8-k} \cdot x^{8-k} \cdot 3^k = \binom{8}{k} \cdot 2^{8-k} \cdot 3^k \cdot x^{8-k}
```
To get $$x^5$$, we need $$8-k = 5$$, so $$k = 3$$.
4. **Calculate carefully**
Substituting $$k = 3$$:
```math
\binom{8}{3} \cdot 2^{8-3} \cdot 3^3 \cdot x^5 = \binom{8}{3} \cdot 2^5 \cdot 3^3 \cdot x^5
```
```math
= \frac{8!}{3! \cdot 5!} \cdot 32 \cdot 27 \cdot x^5
```
```math
= \frac{8 \times 7 \times 6}{3 \times 2 \times 1} \cdot 32 \cdot 27 \cdot x^5
```
```math
= 56 \cdot 32 \cdot 27 \cdot x^5
```
```math
= 1{,}792 \cdot 27 \cdot x^5 = 48{,}384x^5
```
Therefore, the coefficient of $$x^5$$ is $$48{,}384$$.
Visible text: 1. **Identify components**
From , we get:
-
-
-
2. **Determine the type of term**
We are looking for the coefficient of the term containing .
3. **Use the general term formula**
General term:
Expanding the general term:
To get , we need , so .
4. **Calculate carefully**
Substituting :
Therefore, the coefficient of is .
Remember that **every term in the binomial expansion** has a total power equal to the original power, and binomial coefficients are always symmetric: $$\binom{n}{k} = \binom{n}{n-k}$$.
Visible text: Remember that **every term in the binomial expansion** has a total power equal to the original power, and binomial coefficients are always symmetric: .
## Exercises
1. Determine the coefficient of $$x^4$$ in the expansion of $$(2x - 3)^7$$.
2. Calculate the constant term of $$\left(x^2 + \frac{1}{x}\right)^9$$.
3. In the expansion of $$(1 + 2x)^{10}$$, determine the term containing $$x^3$$.
Visible text: 1. Determine the coefficient of in the expansion of .
2. Calculate the constant term of .
3. In the expansion of , determine the term containing .
### Answer Key
1. **Solution:**
Write in binomial form with $$a = 2x$$, $$b = -3$$, and $$n = 7$$.
General term: $$\binom{7}{k} (2x)^{7-k} (-3)^k = \binom{7}{k} \cdot 2^{7-k} \cdot (-3)^k \cdot x^{7-k}$$
For the coefficient of $$x^4$$, we need $$7-k = 4$$, so $$k = 3$$.
```math
\binom{7}{3} \cdot 2^{7-3} \cdot (-3)^3 = \binom{7}{3} \cdot 2^4 \cdot (-3)^3
```
```math
= \frac{7!}{3! \cdot 4!} \cdot 16 \cdot (-27)
```
```math
= \frac{7 \times 6 \times 5}{3 \times 2 \times 1} \cdot 16 \cdot (-27)
```
```math
= 35 \cdot 16 \cdot (-27) = 560 \cdot (-27) = -15{,}120
```
Therefore, the coefficient of $$x^4$$ is $$-15{,}120$$.
2. **Solution:**
Write in binomial form with $$a = x^2$$, $$b = \frac{1}{x}$$, and $$n = 9$$.
```math
\text{General term: } \binom{9}{k} (x^2)^{9-k} \left(\frac{1}{x}\right)^k = \binom{9}{k} \cdot x^{2(9-k)} \cdot x^{-k}
```
```math
= \binom{9}{k} \cdot x^{18-2k} \cdot x^{-k} = \binom{9}{k} \cdot x^{18-2k-k} = \binom{9}{k} \cdot x^{18-3k}
```
For the constant term, the power of $$x$$ must be zero: $$18-3k = 0$$, so $$k = 6$$.
```math
\binom{9}{6} = \binom{9}{3} = \frac{9!}{3! \cdot 6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84
```
Therefore, the constant term is $$84$$.
3. **Solution:**
Write in binomial form with $$a = 1$$, $$b = 2x$$, and $$n = 10$$.
General term: $$\binom{10}{k} (1)^{10-k} (2x)^k = \binom{10}{k} \cdot 2^k \cdot x^k$$
For the term containing $$x^3$$, we need $$k = 3$$.
```math
\binom{10}{3} \cdot 2^3 \cdot x^3 = 120 \cdot 8 \cdot x^3 = 960x^3
```
Therefore, the term containing $$x^3$$ is $$960x^3$$.
Visible text: 1. **Solution:**
Write in binomial form with , , and .
General term:
For the coefficient of , we need , so .
Therefore, the coefficient of is .
2. **Solution:**
Write in binomial form with , , and .
For the constant term, the power of must be zero: , so .
Therefore, the constant term is .
3. **Solution:**
Write in binomial form with , , and .
General term:
For the term containing , we need .
Therefore, the term containing is .