Problem 45 - Algebra and Trigonometry with Analytic Geometry

Question

Without expanding completely, find the indicated term(s) in the expansion of the expression. $$ \left(3 x-\frac{1}{4 x}\right)^{6} ; \quad \text { term that does not contain } x $$

Step-by-Step Solution

Answer

The term without \(x\) is \(-135/8\).

Step 1: Identify the General Term

The general term in the expansion of \((a + b)^n\) is given by \(T_k = \binom{n}{k}a^{n-k}b^k\). For the expression \(\left(3x - \frac{1}{4x}\right)^6\), identify the general term as \(T_k = \binom{6}{k}(3x)^{6-k}\left(-\frac{1}{4x}\right)^k\).

Step 2: Simplify the General Term

Simplify the expression for the general term: \(T_k = \binom{6}{k}(3)^{6-k}x^{6-k}\left(-\frac{1}{4}\right)^k \cdot x^{-k}\). This leads to \(T_k = \binom{6}{k}(3)^{6-k}\left(-\frac{1}{4}\right)^k x^{6-2k}\).

Step 3: Find the Term Without Variable

We need the exponent of \(x\) to be zero (i.e. the term that does not contain \(x\)). Set the equation for the exponent \(6-2k=0\). Solve for \(k\) to find \(k=3\).

Step 4: Calculate the Coefficient for the Specific Term

Substitute \(k=3\) into the general term: \(T_3 = \binom{6}{3}(3)^{6-3}\left(-\frac{1}{4}\right)^3\). Evaluate this expression: \(T_3 = 20 \cdot 27 \cdot \left(-\frac{1}{64}\right)\).

Key Concepts

General Term in Binomial Expansion

The concept of the general term in binomial expansion is crucial in finding specific terms within the expansion of a binomial expression. Binomial expansion is used when you want to expand expressions in the form of \((a + b)^n\). The general formula for any term in the expansion is given by:

• \(T_k = \binom{n}{k} a^{n-k} b^k\)

Here, \(\binom{n}{k}\) is a binomial coefficient, which represents the number of ways to choose \(k\) elements from \(n\) elements. In our problem, the expression is \(\left(3x - \frac{1}{4x}\right)^6\). We can identify the variables as \(a = 3x\) and \(b = -\frac{1}{4x}\), with \(n = 6\). This means that the general term \(T_k\) for any term we wish to find is:

• \(T_k = \binom{6}{k} (3x)^{6-k} \left(-\frac{1}{4x}\right)^k\)

This formula allows you to pick any \(k\)-th term in the expansion.

Exponent Zero Term

To find the term without the variable \(x\), we need the exponent of \(x\) to be zero. This is known as the exponent zero term. In our expression, the general term is:

• \(T_k = \binom{6}{k}(3)^{6-k}x^{6-k}\left(-\frac{1}{4}\right)^k \cdot x^{-k}\)

Simplifying the powers of \(x\), we get:

• \(T_k = \binom{6}{k}(3)^{6-k}\left(-\frac{1}{4}\right)^k x^{6-2k}\)

To find out where the exponent of \(x\) is zero, you set the exponent equal to zero:

• \(6 - 2k = 0\)

Solving this equation, we find that \(k = 3\). Therefore, the term where \(x\) is not present is the 3rd term.

Coefficient Calculation

After identifying the term without \(x\), next is calculating its coefficient. Substituting \(k = 3\) into the general term formula:

• \(T_3 = \binom{6}{3}(3)^{6-3}\left(-\frac{1}{4}\right)^3\)

First, calculate the binomial coefficient \(\binom{6}{3}\), which is 20. Then, calculate \((3)^{6-3}\), which equals 27. Also compute \(\left(-\frac{1}{4}\right)^3\), which equates to \(-\frac{1}{64}\). Multiplying these components gives:

• \(T_3 = 20 \times 27 \times \left(-\frac{1}{64}\right)\)

After evaluating, the coefficient for the term that does not contain \(x\) is \(-\frac{135}{32}\). Thus, this is your desired term without any \(x\) present in its expression.