Problem 30 - Precalculus Mathematics for Calculus

Question

Find the first four terms in the expansion of \(\left(x^{1 / 2}+1\right)^{30}\).

Step-by-Step Solution

Answer

The first four terms are \( x^{15}, 30x^{14.5}, 435x^{14}, \text{ and } 4060x^{13.5} \).

Step 1: Identify the Binomial Theorem

The binomial theorem states that for any positive integer \( n \) and any real number \( a \), \( (x+a)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} a^k \). In this problem, \( x = x^{1/2} \), \( a = 1 \), and \( n = 30 \).

Step 2: Calculate the First Term

The first term in the expansion is obtained where \( k = 0 \). Using the binomial coefficient \( \binom{30}{0} \), we calculate: \( \binom{30}{0} (x^{1/2})^{30-0} 1^0 = x^{15} \). Hence, the first term is \( x^{15} \).

Step 3: Calculate the Second Term

The second term is obtained using \( k = 1 \). Calculate \( \binom{30}{1} (x^{1/2})^{30-1} 1^1 = 30 x^{14.5} \). The second term is \( 30x^{14.5} \).

Step 4: Calculate the Third Term

For the third term, use \( k = 2 \). Calculate \( \binom{30}{2} (x^{1/2})^{30-2} 1^2 = 435 x^{14} \). Thus, the third term is \( 435x^{14} \).

Step 5: Calculate the Fourth Term

For the fourth term, use \( k = 3 \). Calculate \( \binom{30}{3} (x^{1/2})^{30-3} 1^3 = 4060 x^{13.5} \). Thus, the fourth term is \( 4060x^{13.5} \).

Key Concepts

Binomial Expansion

Binomial expansion is a way to express the power of a sum, such as \((x+a)^n\), in terms of its individual terms. This technique is crucial because it breaks down complex expressions into simpler components. In our current example, we are expanding \((x^{1/2} + 1)^{30}\). Understanding binomial expansion allows us to find the terms efficiently. Each term in the expansion represents a combination of powers of \(x\) and \(a\). These terms are linked by plus signs, forming a polynomial of degree \(n\). It is a method that effectively uses patterns and simplifications to derive results that might otherwise require tedious calculation.One key aspect of binomial expansion is that it contributes significantly to simplifying the problem into smaller, easier parts. By calculating only the first few terms, one can approximate or partially understand the complete expansion.

Binomial Coefficients

Binomial coefficients play a vital role in the binomial expansion process. They determine the weight or the frequency of each term in the expansion. These coefficients are represented by \(\binom{n}{k}\), pronounced as "n choose k". In mathematics, these coefficients are the number of ways to choose \(k\) elements from a set of \(n\) elements, irrespective of the order. They are calculated using the formula:

• \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\)

For instance, in our specific problem, to find the coefficient for each term in the expansion \((x^{1/2} + 1)^{30}\), we utilize these binomial coefficients: \(\binom{30}{0}\), \(\binom{30}{1}\), \(\binom{30}{2}\), and \(\binom{30}{3}\). These coefficients influence the polynomial outcome by scaling the base terms accordingly.

Exponentiation

Exponentiation involves raising numbers to a power. This is a fundamental operation in mathematics and plays a crucial role in generating terms within a binomial expansion. For the expansion of \((x^{1/2} + 1)^{30}\), exponentiation helps in determining the powers of \(x\) in each term. Since the base involves \(x^{1/2}\), each term is affected by this initial exponent. For example:

• The first term \((x^{1/2})^{30-0}\) simplifies to \(x^{15}\).
• The second term \((x^{1/2})^{30-1}\) simplifies to \(x^{14.5}\).

Exponentiation allows us to handle non-integer exponents fluently, providing the essential base changes in each calculated term. Understanding this process aids in rounding off decimal exponents when needed, contributing to clearer solutions.

Polynomial Expansion

Polynomial expansion is the result of applying the binomial theorem, generating a series of terms. Each term consists of variables raised to various powers, often with coefficients. In our example, the binomial expansion leads to a polynomial that starts with \(x^{15}\), followed by \(30x^{14.5}\), \(435x^{14}\), and \(4060x^{13.5}\). The benefit of polynomial expansion is that it decomposes a complex power expression into simpler, manageable sums of terms. This expansion method allows us to approximate functions or expressions, predict and verify results, and solve equations analytically. It applies heavily in calculus, statistics, and numerical analysis, where breaking down functions into polynomial components can simplify complex problems. Thus, mastering polynomial expansion can enhance problem-solving skills significantly.