Binomial Newton

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.

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:

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:

Where $\binom{n}{k}$ is the binomial coefficient calculated with the formula:

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:

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 $n$.

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 $(1 - x)^{2014}$.

First, we rewrite it in standard binomial form with $a = 1$, $b = -x$, and $n = 2014$:

To get the term containing $x^2$, we need $k = 2$:

Calculating the binomial coefficient:

Therefore, the coefficient of $x^2$ is $2{,}026{,}591$.

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 $\left(3x^3 - \frac{2}{x}\right)^8$.

We write it in binomial form with $a = 3x^3$ and $b = -\frac{2}{x}$:

The general term is:

For the constant term, the power of $x$ must be zero:

Substituting $k = 6$:

Note that $(-2)^6 = 64$ because even powers always produce positive values, just like $(+2)^6 = 64$.

Therefore, the constant term is $16{,}128$.

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.

Example Strategy Application:

Determine the coefficient of $x^5$ in the expansion of $(2x + 3)^8$.

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:

   $$\binom{8}{k} (2x)^{8-k} (3)^k = \binom{8}{k} \cdot 2^{8-k} \cdot x^{8-k} \cdot 3^k = \binom{8}{k} \cdot 2^{8-k} \cdot 3^k \cdot x^{8-k}$$

   To get $x^5$, we need $8-k = 5$, so $k = 3$.

4. Calculate carefully

   Substituting $k = 3$:

   $$\binom{8}{3} \cdot 2^{8-3} \cdot 3^3 \cdot x^5 = \binom{8}{3} \cdot 2^5 \cdot 3^3 \cdot x^5$$

   $$= \frac{8!}{3! \cdot 5!} \cdot 32 \cdot 27 \cdot x^5$$

   $$= \frac{8 \times 7 \times 6}{3 \times 2 \times 1} \cdot 32 \cdot 27 \cdot x^5$$

   $$= 56 \cdot 32 \cdot 27 \cdot x^5$$

   $$= 1{,}792 \cdot 27 \cdot x^5 = 48{,}384x^5$$

   Therefore, the coefficient of $x^5$ is $48{,}384$.

Remember that every term in the binomial expansion has a total power equal to the original power, and binomial coefficients are always symmetric: $\binom{n}{k} = \binom{n}{n-k}$.

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 $x^3$.

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 $k = 3$.

   $$\binom{7}{3} \cdot 2^{7-3} \cdot (-3)^3 = \binom{7}{3} \cdot 2^4 \cdot (-3)^3$$

   $$= \frac{7!}{3! \cdot 4!} \cdot 16 \cdot (-27)$$

   $$= \frac{7 \times 6 \times 5}{3 \times 2 \times 1} \cdot 16 \cdot (-27)$$

   $$= 35 \cdot 16 \cdot (-27) = 560 \cdot (-27) = -15{,}120$$

   Therefore, the coefficient of $x^4$ is $-15{,}120$.

2. Solution:

   Write in binomial form with $a = x^2$, $b = \frac{1}{x}$, and $n = 9$.

   $$\text{General term: } \binom{9}{k} (x^2)^{9-k} \left(\frac{1}{x}\right)^k = \binom{9}{k} \cdot x^{2(9-k)} \cdot x^{-k}$$

   $$= \binom{9}{k} \cdot x^{18-2k} \cdot x^{-k} = \binom{9}{k} \cdot x^{18-2k-k} = \binom{9}{k} \cdot x^{18-3k}$$

   For the constant term, the power of $x$ must be zero: $18-3k = 0$, so $k = 6$.

   $$\binom{9}{6} = \binom{9}{3} = \frac{9!}{3! \cdot 6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$$

   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 $k = 3$.

   $$\binom{10}{3} \cdot 2^3 \cdot x^3 = 120 \cdot 8 \cdot x^3 = 960x^3$$

   Therefore, the term containing $x^3$ is $960x^3$.