Binomial Newton

How Binomial Coefficients Change with the Power

Expanding $(x + y)^{10}$ by repeated multiplication requires many steps. Newton's binomial theorem writes $(x + y)^n$ directly as a sum whose coefficients are binomial coefficients.

The coefficients follow a predictable pattern. Once we understand that pattern, the same formula works for every nonnegative integer power.

Look at the basic patterns for the first few powers:

The coefficients change with the power, but their positions follow one consistent rule.

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 read as "$n$ choose $k$" because it counts the 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

Newton's binomial theorem can find the coefficient of a selected term without expanding the entire expression.

Suppose we want to find the coefficient of $x^2$ in the expansion of $(1 - x)^{2014}$.

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:

The coefficient of $x^2$ is $2{,}027{,}091$.

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 $k = 6$:

Both $(-2)^6$ and $(+2)^6$ equal $64$ because an even power of a nonzero real number is positive.

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:

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

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

4. Calculate carefully

   Substituting $k = 3$:

   $$\begin{aligned} &\binom{8}{3} \cdot 2^{8-3} \cdot 3^3 \cdot x^5 \\ &= \binom{8}{3} \cdot 2^5 \cdot 3^3 \cdot x^5 \end{aligned}$$

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

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

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

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

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

Two checks help catch mistakes: every term in the expansion has total degree $n$, and the binomial coefficients are symmetric: $\binom{n}{k} = \binom{n}{n-k}$.

Exercises

Each problem asks for a single term of one expansion. Use the general term formula and match the exponent that the question names.

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 $k = 3$.

   $$\begin{aligned} &\binom{7}{3} \cdot 2^{7-3} \cdot (-3)^3 \\ &= \binom{7}{3} \cdot 2^4 \cdot (-3)^3 \end{aligned}$$

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

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

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

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

2. Solution:

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

   The general term is:

   $$\begin{aligned} &\binom{9}{k} (x^2)^{9-k} \left(\frac{1}{x}\right)^k \\ &=\binom{9}{k} \cdot x^{2(9-k)} \cdot x^{-k} \end{aligned}$$

   $$\begin{aligned} &= \binom{9}{k} \cdot x^{18-2k} \cdot x^{-k} \\ &= \binom{9}{k} \cdot x^{18-2k-k} \\ &= \binom{9}{k} \cdot x^{18-3k} \end{aligned}$$

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

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

   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 $k = 3$.

   $$\begin{aligned} \binom{10}{3} \cdot 2^3 \cdot x^3 &= 120 \cdot 8 \cdot x^3 \\ &= 960x^3 \end{aligned}$$

   The term containing $x^3$ is $960x^3$.