Problem 39 - Precalculus: Mathematics for Calculus

Question

Find the indicated terms in the expansion of the given binomial. The term containing \(x^{4}\) in the expansion of \((x+2 y)^{10}\).

Step-by-Step Solution

Answer

The term containing \(x^4\) is \(13440x^4y^6\).

Step 1: Identify the Binomial Formula

To find a specific term in a binomial expansion, use the general term formula for a binomial. The general term for any binomial expansion \((x + a)^n\) is given by \( T_{k+1} = \binom{n}{k} x^{n-k} a^k \). In this case, the binomial is \((x + 2y)^{10}\).

Step 2: Determine the General Term

Using the formula from Step 1, the general term \(T_{k+1}\) can be written as \(T_{k+1} = \binom{10}{k} x^{10-k} (2y)^k\). Simplify this to get the terms in terms of \(x\) and \(y\): \(T_{k+1} = \binom{10}{k} x^{10-k} \cdot 2^k \cdot y^k\).

Step 3: Find the Exponent of x

We need to find the term where the power of \(x\) is 4. From the general term \(x^{10-k}\), set \(10-k = 4\). Solving for \(k\) gives \(k = 6\).

Step 4: Substitute k into the General Term

Substitute \(k = 6\) into the equation from Step 2 to find the specific term: \(T_{7} = \binom{10}{6} x^{4} (2y)^6\).

Step 5: Calculate the Binomial Coefficient and Simplify

Calculate the binomial coefficient: \(\binom{10}{6} = 210\). Then substitute to get \(T_7 = 210 \cdot x^4 \cdot 2^6 \cdot y^6\). Simplify \(2^6 = 64\) so the term is \(210 \cdot 64 \cdot x^4 \cdot y^6\).

Step 6: Calculate Final Term

Multiply the coefficients: \(210 \times 64 = 13440\). Therefore, the term is \(13440x^4y^6\).

Key Concepts

Binomial Expansion

The concept of binomial expansion is an essential part of algebra and helps us efficiently expand expressions raised to a power. Imagine expanding something like \((x + y)^n\). Instead of multiplying \((x + y)\) by itself \(n\) times, the binomial theorem gives us a shortcut.By using binomial expansion, we can calculate each term without having to do extensive multiplication. The binomial theorem states that for any integer \(n\):\[(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k\]This formula provides all the terms in the expansion, where \(\binom{n}{k}\) is a binomial coefficient, representing the ways to choose \(k\) elements from a set of \(n\) elements.
Understanding binomial expansion simplifies working with polynomials and is widely applied in calculus and probability theory.

General Term Formula

The general term formula is the key to unlocking specific terms in a binomial expansion. It helps in directly finding any term without needing the entire expansion.
For a binomial raised to the power of \(n\), like \((x + a)^n\), we use:\[ T_{k+1} = \binom{n}{k} x^{n-k} a^k \]This formula determines the \((k+1)\)th term of the expansion. Each component in the formula serves a specific purpose:

• The binomial coefficient \(\binom{n}{k}\) calculates how many ways the terms can be combined.
• The component \(x^{n-k}\) dictates the power of \(x\).
• The part \(a^k\) informs us about the power of the second term.

By substituting different values of \(k\), you find each term in the expansion. As seen in the exercise, finding terms like the one containing \(x^4\) just needs solving for \(k\) in the respective the equation.

Exponents in Binomials

Exponents in binomial expansions play a crucial role in determining the power of each part of the expanded expression. When expanding, the coefficients and terms are decided by both the exponents and the binomial coefficients.
Each term in a binomial expansion \((x + a)^n\) will have a format that follows \(x^{n-k}\) and \(a^k\), indicating:

• The first part of the term involves the decreasing power of \(x\), as seen with \(x^{n-k}\).
• The second part involves the increasing power of \(a\), represented as \(a^k\).

In our specific example, we needed the term with \(x^4\). This means finding the value of the exponent that satisfies the condition for \(x\).
By setting \(n-k=4\), we identified \(k\) to be \(6\) in the expansion of \((x + 2y)^{10}\).
Understanding how to manipulate these exponents is vital for solving problems efficiently, as it aids in pinpointing specific terms without confusion.