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Chapter 10: Problem 38

Find the coefficient \(C\) of the given term in each binomial expansion. Binomial Term $$(3+y)^{9} \quad c y^{5}$$

Short Answer

Expert verified
The coefficient \(C\) for the term \(y^5\) in \((3+y)^9\) is 10206.

Step by step solution

01

Identify the Binomial Expansion Formula

The binomial expansion of \((a+b)^n\) is given by \(\sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). We need to use this formula to expand \((3+y)^9\).
02

Match the Term to the Formula

We need to find the coefficient of the term which includes \(y^5\) in the expansion of \((3+y)^9\). In the formula, \(b = y\) and the power \(k = 5\).
03

Calculate the Binomial Coefficient

The binomial coefficient can be calculated as \(\binom{9}{5}\). This represents the number of ways to choose 5 elements from a set of 9, and is calculated as \(\frac{9!}{5!(9-5)!}\).
04

Simplify the Binomial Coefficient

Calculate \(\binom{9}{5} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126\).
05

Determine the Coefficient from the Expanded Term

In the term \(\binom{9}{5} a^{9-5} b^5\), substitute \(a = 3\), \(b = y\). This gives us \(126 \times 3^{4} \times y^5\).
06

Calculate the Final Coefficient

Calculate \(3^4 = 81\). The coefficient \(C\) is thus \(126 \times 81 = 10206\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binomial Coefficient
The binomial coefficient is a crucial concept in the world of algebra, especially when dealing with expressions that require expansion. It essentially counts how many ways you can choose \( k \) items from a larger set of \( n \) items, represented as \[ \binom{n}{k} \]. This concept is mathematically expressed as \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \], where \( ! \) denotes a factorial, which is the product of all positive integers up to a certain number.
  • For example, finding \( \binom{9}{5} \) means calculating the ways to choose 5 items from a total of 9, and is computed as \[ \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \].
  • This coefficient is not merely a mathematical curiosity; it is the weight or multiplier for each term in the expansion of a binomial expression.
Binomial Theorem
The Binomial Theorem is a powerful algebraic tool that describes how to expand expressions that are raised to a power, such as \( (a+b)^n \). This theorem provides a straightforward way to expand polynomials without manually multiplying the binomials.According to the Binomial Theorem, any binomial expression \( (a+b)^n \) can be expanded using the formula \[ \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \].
  • Each term in this expansion is multiplied by a binomial coefficient \( \binom{n}{k} \), which we learned counts the number of ways to choose items in combinations, making it central to finding term coefficients in binomial expansions.
  • This formula efficiently helps list out each term in the expansion, calculated for specified powers in descending order of \( a \) and ascending order of \( b \).
Once understood, this method significantly simplifies our work in algebra, especially for polynomial expansions where more than two terms are involved.
Polynomial Expansion
When expanding polynomials using the Binomial Theorem, you are potentially simplifying a complex expression into a series of terms. Each term results from specifying values of \( a \), \( b \), and \( n \) from our binomial expression.In a typical polynomial expansion of \( (3+y)^9 \), for instance, we use the theorem to break down and redefine its terms based on powers of \( y \) and constants multiplied by coefficients.
  • The term connected to \( y^5 \) involves determining its coefficient through \( \binom{9}{5} \), substituting the value of \( a \) (in our example, 3), raising it to the appropriate power, and then multiplying by \( b^5 \).
  • Using these steps, we find specific terms like 126 for the coefficient, resulting in the expression \( 126 \times 3^4 \times y^5 \), and at the end, evaluating 3 to the power of 4 results in 81, which ultimately gives us a total coefficient of 10,206 when calculated thoroughly.
This approach allows you to grasp each part of the expansion, ensuring that every detail is considered for accurate results.

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