Problem 38 - Precalculus: Mathematical for Calculus

Question

Find the term that does not contain \(x\) in the expansion of $$ \left(8 x+\frac{1}{2 x}\right)^{8} $$

Step-by-Step Solution

Answer

The term independent of \(x\) is 17920.

Step 1: Understand the Binomial Theorem

The binomial theorem states that for any positive integer \(n\), \((a + b)^n\) expands as \(\sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). Here, \(a = 8x\) and \(b = \frac{1}{2x}\). We need the expansion of \((8x + \frac{1}{2x})^8\).

Step 2: Identify the General Term

The general term in the expansion is given by \( T_k = \binom{8}{k} (8x)^{8-k} \left(\frac{1}{2x}\right)^k\). Simplify the general term: \[ T_k = \binom{8}{k} \cdot 8^{8-k} \cdot x^{8-k} \cdot \left(\frac{1}{2}\right)^k \cdot x^{-k} \] Thus, \[ T_k = \binom{8}{k} \cdot 8^{8-k} \cdot \left(\frac{1}{2}\right)^k \cdot x^{8-2k} \].

Step 3: Find the Term Independent of x

To find the term independent of \(x\), set the power of \(x\) to zero:\[ 8 - 2k = 0 \]Solve for \(k\): \[ 2k = 8 \rightarrow k = 4 \].

Step 4: Calculate the Specific Term

Substitute \(k = 4\) into the general term to compute the term without \(x\):\[ T_4 = \binom{8}{4} \cdot 8^{8-4} \cdot \left(\frac{1}{2}\right)^4 \cdot x^{0} \]Calculate \(\binom{8}{4}=70\), thus,\[ T_4 = 70 \cdot 8^4 \cdot \left(\frac{1}{2}\right)^4 \].

Step 5: Simplify the Expression

Calculate \(8^4 = 4096\) and \(\left(\frac{1}{2}\right)^4 = \frac{1}{16}\). Therefore, \[ T_4 = 70 \cdot 4096 \cdot \frac{1}{16} \]Calculate the final expression:\[ T_4 = 70 \cdot 256 \]\[ T_4 = 17920 \].

Key Concepts

Binomial Expansion

The binomial expansion is a fundamental concept in algebra, used to expand expressions raised to a power. For the expression \( (a + b)^n \), the Binomial Theorem helps us expand it in terms of its components. This theorem is particularly useful for managing expressions where directly computing powers is cumbersome.

• It states that: \[ (a + b)^n = inom{n}{0} a^n b^0 + inom{n}{1} a^{n-1} b^1 + inom{n}{2} a^{n-2} b^2 + ext{...} + inom{n}{n} a^0 b^n \]
• Each term in the expansion involves coefficients known as binomial coefficients, represented by \(\binom{n}{k}\), which indicate the number of ways to choose \(k\) elements from \(n\) elements.
• The binomial coefficients can be easily found by using Pascal's Triangle or directly calculating with the formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \).
• In our exercise, \( a = 8x \) and \( b = \frac{1}{2x} \), and we are interested in expanding \( (8x + \frac{1}{2x})^8 \).

General Term

The general term in a binomial expansion allows us to focus on specific terms without writing out the entire series. This is crucial when searching for particular terms in an expansion, like those that simplify nicely or are independent of a variable.

• The general term \( T_k \) in the binomial expansion \((a + b)^n\) is given by \[ T_k = \binom{n}{k} a^{n-k} b^k \]
• In solving our problem, the expansion \((8x + \frac{1}{2x})^8\) uses the general term formula to find \( T_k = \binom{8}{k} (8x)^{8-k} (\frac{1}{2x})^k \).
• This was further simplified to \[ T_k = \binom{8}{k} \cdot 8^{8-k} \cdot \left(\frac{1}{2}\right)^k \cdot x^{8-2k} \], highlighting how the variable \(x\) interacts with the terms.

Instead of expanding all terms, we efficiently find the one that does not contain \( x \), which requires the power of \( x \) in the term to be zero.

Power of x

Understanding the power of \( x \) in each term is key when determining specific components of a binomial expansion. It describes how \( x \) is manipulated and what its influence on each term is.

• The expression for our exercise simplifies to terms of the form \( x^{8-2k} \), where each term’s power of \( x \) is \( 8-2k \).
• To find a term independent of \( x \), set \( 8-2k = 0 \). Solving this equation gives \( k = 4 \).
• Thus, substituting \( k = 4 \) in the general term confirms this specific term will eliminate \( x \) wholly, leading us to the terms that don't contain \( x \).

Here, finding \( k = 4 \) was the pivotal solution step, allowing us to isolate the desired term independent of any variables.