Problem 42 - Precalculus Mathematics 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 constant term is 17920.

Step 1: Identify the General Term

In the expansion of \((a+b)^n\), the general term is given by \(T_{k+1} = \binom{n}{k} a^{n-k} b^k\). For \(\left(8x + \frac{1}{2x}\right)^8\), we have \(a = 8x\) and \(b = \frac{1}{2x}\), with \(n = 8\). The general term becomes \(T_{k+1} = \binom{8}{k} (8x)^{8-k} \left(\frac{1}{2x}\right)^k\).

Step 2: Simplify the General Term

Simplify \(T_{k+1} = \binom{8}{k} (8x)^{8-k} \left(\frac{1}{2x}\right)^k\). This becomes \(T_{k+1} = \binom{8}{k} \cdot 8^{8-k} \cdot x^{8-k} \cdot \frac{1}{2^k} \cdot x^{-k}\). Simplifying further, we get \(T_{k+1} = \binom{8}{k} \cdot \frac{8^{8-k}}{2^k} \cdot x^{8-2k}\).

Step 3: Find the Constant Term

The coefficient of the term that does not contain \(x\) (constant term) occurs when the power of \(x\) is zero. Set the power of \(x\) in the simplified general term to zero: \(8-2k=0\). Solving this equation gives \(k = 4\).

Step 4: Calculate the Constant Term

Substitute \(k=4\) into the expression \(T_{k+1} = \binom{8}{k} \cdot \frac{8^{8-k}}{2^k}\). We calculate \(\binom{8}{4} = 70\), \(8^{8-4} = 8^4 = 4096\), and \(2^4 = 16\). Substituting these values gives \(T_{5} = 70 \cdot \frac{4096}{16} = 70 \cdot 256 = 17920\).

Key Concepts

General Term

The general term in a binomial expansion is a crucial concept, often represented as \(T_{k+1} = \binom{n}{k} a^{n-k} b^k\). This formula helps us find any term in the expansion of \((a+b)^n\). Here, \(n\) is the total number of terms minus one, \(a\) and \(b\) are the variables or constants being expanded, and \(k\) is a specific term number.

In the exercise provided, the expression \((8x + \frac{1}{2x})^8\) involves expanding into individual terms. Our general term here simplifies to \(T_{k+1} = \binom{8}{k} (8x)^{8-k} \left(\frac{1}{2x}\right)^k\). Each component of this equation plays a role:

• \(\binom{8}{k}\) represents the binomial coefficient, providing the number of ways to choose \(k\) elements from \(8\).
• \((8x)^{8-k}\) and \(\left(\frac{1}{2x}\right)^k\) involve raising \(8x\) and \(\frac{1}{2x}\) to particular powers, helping to formulate the term.

Simplifying the equation gives us insight into how terms behave when expanded. For each \(k\), we compute a different term; the combination of these forms the expansion.

Constant Term

In binomial expansion, a constant term is one without variables, in this case, without \(x\). To find this, our goal is to set the power of \(x\) to zero.

Through simplification, the task is achieved by finding when the exponents of all \(x\) terms cancel out. For the exercise, the power of \(x\) winds down through \(8-2k\). When we place this equal to zero, we solve for \(k\):

• \(8 - 2k = 0\)
• Solving gives \(k = 4\).

With \(k = 4\), we substitute back into our general term to find the constant term. This method reveals which term in the expansion is a numerical constant, a key to resolving problems involving variable-free components.

Binomial Coefficients

Binomial coefficients such as \(\binom{n}{k}\) are parts of the binomial theorem. These coefficients show how many ways we can choose \(k\) elements from \(n\) elements, often read as "n choose k." In the binomial expansion, these coefficients dictate the weight or multiplicity of each term.

For the exercise, the coefficient \(\binom{8}{k}\) defines each term contribution. Computed through \(\binom{8}{4} = 70\), it is a critical component in balancing the expression.

• These values rely on factorial calculation: \(\binom{n}{k} = \frac{n!}{k! (n-k)!}\).
• This dictates the amplification or reduction of each term generated.

Understanding binomial coefficients helps grasp the structure and magnitude of polynomial terms during expansion.