Problem 38 - Precalculus: Mathematics for Calculus

Question

Find the indicated terms in the expansion of the given binomial. The second term in the expansion of \(\left(x^{2}-\frac{1}{x}\right)^{25}\).

Step-by-Step Solution

Answer

The second term is \(-25x^{47}\).

Step 1: Identify the Binomial Expansion Formula

The binomial theorem states that \((a + b)^{n} = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}\). We will apply this formula to \((x^{2} - \frac{1}{x})^{25}\).

Step 2: Apply the Formula

In this binomial \(a = x^{2}\) and \(b = -\frac{1}{x}\), with \(n = 25\). The general term \(T_{k+1}\) in the expansion can be given by the formula: \(T_{k+1} = \binom{n}{k} (x^{2})^{n-k} \left(-\frac{1}{x}\right)^{k}\).

Step 3: Find the Second Term

For the second term of the expansion, the value of \(k\) is 1. Substitute \(k = 1\) into the general term formula: \(T_{2} = \binom{25}{1} (x^{2})^{25-1} \left(-\frac{1}{x}\right)^{1}\).

Step 4: Simplify the Expression

Calculate each part of the formula:- \(\binom{25}{1} = 25\)- \((x^{2})^{24} = x^{48}\)- \(\left(-\frac{1}{x}\right)^{1} = -\frac{1}{x}\)Putting it all together: \(T_{2} = 25 \cdot x^{48} \cdot (-\frac{1}{x})\).

Step 5: Compute the Final Expression

Now multiply the components: \(T_{2} = 25 \cdot x^{48} \cdot -\frac{1}{x} = -25 \cdot x^{47}\). This is the expression for the second term.

Key Concepts

Binomial Expansion

The binomial theorem provides a way to expand expressions that are raised to a power, such as \((x^2 - \frac{1}{x})^{25}\). The key formula here is\((a + b)^{n} = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}\).
This formula is incredibly useful because it breaks down complex expressions into more manageable terms.

Each term in the expansion of a binomial expression is formed by taking the coefficients, known as the binomial coefficients, from Pascal's Triangle or calculated using the combination formula \(\binom{n}{k}\).
Here, \(n\) is the power the binomial is raised to, and \(k\) is the specific term number minus one since counting starts from zero.

For the expression \((x^2 - \frac{1}{x})^{25}\), we see that the pattern follows immediately when we apply the binomial theorem:

• \(a = x^2\) is the first term of the binomial.
• \(b = -\frac{1}{x}\) is the second term of the binomial.

Breaking down the problem using the binomial theorem allows for straightforward identification of any specific term in the expansion.

General Term Formula

The general term formula in the binomial expansion gives us the ability to find any term in an expanded binomial expression directly.
I in the context of \((x^2 - \frac{1}{x})^{25}\), the formula takes the form:\[T_{k+1} = \binom{n}{k} (x^2)^{n-k} \left(-\frac{1}{x}\right)^{k}\]
This formula allows us to identify each term by calculating the specific powers of \(x^2\) and \(-\frac{1}{x}\) and combining them with the respective binomial coefficients.

In practice, finding the second term involves setting \(k=1\), as the first term corresponds to \(k=0\). Thus, substituting \(k=1\) into the general term formula, we derive:

• \(\binom{25}{1}\) calculates the coefficient, yielding 25.
• \((x^2)^{24}\) becomes \(x^{48}\).
• \((-\frac{1}{x})^{1}\) simplifies to \(-\frac{1}{x}\).

Each of these parts contributes to forming the second term after combining them.

Combining Like Terms

Combining like terms is an essential skill in simplifying algebraic expressions, including those from binomial expansions.
In this context, after determining the second term with the formula,\(T_{2} = 25 \cdot x^{48} \cdot (-\frac{1}{x})\), we calculate the product.

This step involves multiplication, where care must be taken to carefully apply rules of algebra.

• The constant multiple \(25\) remains unchanged.
• The expression \(x^{48} \cdot (-\frac{1}{x})\) requires powers of \(x\) to be combined.
• Simplify \(x^{48} \cdot \frac{1}{x} = x^{48-1} = x^{47}\).

The final expression for the second term becomes \(-25x^{47}\). It’s crucial to manage the sign and powers accurately.

Combining like terms simplifies the expression and ensures that each term in our expansion is clear and concise, ready for further use or analysis in mathematical problems.