Problem 48 - A Graphical Approach to Precalculus with Limits

Question

Work each of the following. Find the term in the expansion of \((3+\sqrt{x})^{11}\) that con\(\operatorname{tains} x^{4}\)

Step-by-Step Solution

Answer

The term containing \(x^4\) is \(4455x^4\).

Step 1: Identify the formula for the binomial expansion

To find a specific term in the binomial expansion, use the general term formula in the expansion of \[(a + b)^n\] which is given by \[T_{k+1} = \binom{n}{k} a^{n-k} b^k\]Here, \(a = 3\), \(b = \sqrt{x}\), and \(n = 11\).

Step 2: Set up the conditions for term with x^4

We need to find the term involving \(x^4\). In the general term formula, \(b^k = (\sqrt{x})^k = x^{k/2}\). For the power of \(x\) to be 4, we require: \[\frac{k}{2} = 4\] which gives \(k = 8\).

Step 3: Calculate the coefficient for the term

Now substitute \(k = 8\), \(n = 11\), \(a = 3\), and \(b = \sqrt{x}\) into \[T_{k+1} = \binom{11}{8} 3^{11-8} (\sqrt{x})^8\]Calculate \(\binom{11}{8}\) which is the same as \(\binom{11}{3}\), since \(\binom{n}{k} = \binom{n}{n-k}\). Then, evaluate the expression.

Step 4: Solve for the Term

Using the previous step:\[T_{9} = \binom{11}{8} 3^3 x^4\]Calculate \(\binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\) and \(3^3 = 27\). Therefore:\[T_{9} = 165 \times 27 \times x^4 = 4455 x^4\].

Key Concepts

Power of a Term

In the binomial expansion, when dealing with expressions like \((a + b)^n\), a core concept is understanding the 'power of a term'.
When binomials are expanded, they form series with several terms, each containing distinct powers of the individual variables.
For example, in our case with \((3 + \sqrt{x})^{11}\), the terms will involve powers of 3 and \(\sqrt{x}\) in a specific pattern.
A crucial point is identifying the power of a term, especially when focusing on one particular component—like the power of \(x\) we are interested in (e.g., \(x^4\)).

• The power of \(a\) remains from \(n-k\) where \(a\) is that specific base in the expansion.
• The goal is to determine the appropriate power of the second base, \(b\), that produces the desired exponent in \(x\).

For the expansion \((3 + \sqrt{x})^{11}\), to find \(x^4\), we set the conditions to match its power.
Since \(b^k = (\sqrt{x})^k = x^{k/2}\), our task is to find the required \(k\) making \(x\) equal to the needed degree (in this instance, 4).
Hence, the equation \(\frac{k}{2} = 4\) is solved to \(k = 8\).

General Term Formula

The general term formula is a powerful tool for finding specific terms in a binomial expansion.
It is expressed as \(T_{k+1} = \binom{n}{k} a^{n-k} b^k\).
This formula allows you to calculate any term in the binomial series without expanding the entire expression.
To use it effectively:

• Identify \(n\), \(a\), and \(b\) from your binomial equation.
• Determine the correct \(k\) that aligns with your desired term—here, we're interested in making the term \(x^4\) appear.

After setting the condition for the power of the \(x\) term, solve for \(k\).
In our exercise, \((3 + \sqrt{x})^{11}\), we found \(k = 8\).
Substitute these values into the formula:
\[T_{k+1} = \binom{11}{8} 3^{11-8} (\sqrt{x})^8\].
Calculating these values gives the term that includes \(x^4\).

Combinatorial Coefficients

Combinatorial coefficients, also known as binomial coefficients, represent the number of ways to select \(k\) items from \(n\) items without considering the order.
They are key components of the general term formula.
The notation \(\binom{n}{k}\) is symbolic of these coefficients, which are computed using the formula:\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\].
This provides the weight or multiplier for each term in the expansion.
In our exercise finding the term in \((3 + \sqrt{x})^{11}\) that contains \(x^4\), we use \(\binom{11}{8}\).
Let's simplify it:

• Recognize that \(\binom{11}{8}\) is the same as \(\binom{11}{3}\) because \(\binom{n}{k} = \binom{n}{n-k}\).
• Calculate it: \(\frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\).

These coefficients determine the numerical multiplier in front of each term, as seen in our final term with \(x^4\), which results in \(4455 x^4\).