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Find the term indicated in each expansion. $$\left(x^{2}+y^{3}\right)^{8} ; \text { sixth term }$$

Short Answer

Expert verified
The sixth term of the expansion \((x^{2}+y^{3})^{8}\) is \(56x^{6}y^{15}\)

Step by step solution

01

Identify the binomial theorem

The binomial theorem states that \((a+b)^{n} = \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^{k}\). This allows any term of the expansion to be found directly.
02

Apply the binomial theorem to the specific problem

According to the binomial theorem, the rth term of the expression \((x^{2}+y^{3})^{8}\) can be expressed as \(T_{r} = \binom{8}{r-1} (x^{2})^{8-(r-1)}(y^{3})^{r-1}\), where \(T\) is the rth term. In the problem, the term sought is the 6th term, so \(r\) is 6.
03

Calculate term T6

Substitute \(r = 6\) into the term equation: \(T_{6} = \binom{8}{6-1} (x^{2})^{8-(6-1)}(y^{3})^{6-1} = \binom{8}{5} (x^{2})^{3}(y^{3})^{5} = 56x^{6}y^{15}\).

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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 fundamental concept in combinatorics and plays a central role in the binomial theorem expansion. It is typically denoted as \( \binom{n}{k} \), pronounced as 'n choose k', and represents the number of ways to choose k elements from a set of n distinct elements. In the context of the binomial theorem, it determines the coefficient for each term in the expansion.

Let's break down the math behind it. The general formula for a binomial coefficient is: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] where \( n! \) (n factorial) is the product of all positive integers up to n, and \( k! \) and \( (n-k)! \) are the factorials of k and n-k, respectively. It's important to remember that by definition, \( 0! = 1 \).

  • The value of \( \binom{n}{k} \) is symmetric, meaning \( \binom{n}{k} = \binom{n}{n-k} \).
  • When calculating, if k is larger than n, \( \binom{n}{k} \) is zero since you cannot choose more elements than you have.
  • If k is zero, the coefficient is always 1, mirroring the fact that there is exactly one way to choose nothing from a set.
Binomial Expansion
The binomial expansion is a method to express the power of a binomial, \( (a + b)^n \), as a sum of terms involving binomial coefficients. This concept is vitally useful in algebra and allows us to expand expressions without actually multiplying the binomial by itself n times.

The binomial theorem provides us with a concise formula for these expansions: \((a+b)^{n} = \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^{k}\). Here, \( \binom{n}{k} \) is the binomial coefficient, \( a^{n-k} \) and \( b^{k} \) are the terms of the binomial raised to the appropriate powers, and \( \sum \) denotes the summation of all terms from \( k=0 \) to \( k=n \).

  • Each term in the expansion has a decreasing power of a and an increasing power of b.
  • The coefficients of these terms can be found in Pascal's Triangle or calculated using the binomial coefficient formula.
  • The exponents for a and b in each term always add up to n.
Polynomial Term Calculation
When dealing with binomial expansions, calculating a specific polynomial term efficiently can save a great deal of time and effort. This process, known as polynomial term calculation, involves identifying the term's place in the sequence, establishing the relevant powers for the variables and calculating the binomial coefficient for that term.

Let's consider the term in the exercise. A step-by-step approach was used: identify the term's rank (the sixth term in this case), apply the binomial theorem to find the general form of the term, and then substitute the specific values into the term equation to get the result.

For the given exercise, the calculation was \(T_{6} = \binom{8}{5} (x^{2})^{3}(y^{3})^{5} = 56x^{6}y^{15}\). This illustrates the insightful process of polynomial term calculation by breaking the problem into manageable steps. Specifically:

  • Identifying the rank of the term gives us a clear target within the expansion.
  • Understanding how to apply the binomial theorem helps establish a formula for the term.
  • Calculating the binomial coefficient and determining powers for both variables in the term ensures an accurate result.

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Most popular questions from this chapter

Research and present a group report on state lotteries. Include answers to some or all of the following questions: Which states do not have lotteries? Why not? How much is spent per capita on lotteries? What are some of the lottery games? What is the probability of winning top prize in these games? What income groups spend the greatest amount of money on lotteries? If your state has a lottery, what does it do with the money it makes? Is the way the money is spent what was promised when the lottery first began?

You are now 25 years old and would like to retire at age 55 with a retirement fund of \(\$ 1,000,000 .\) How much should you deposit at the end of each month for the next 30 years in an IRA paying \(10 \%\) annual interest compounded monthly to achieve your goal? Round up to the nearest dollar.

Will help you prepare for the material covered in the next section. The figure shows that when a die is rolled, there are six equally likely outcomes: \(I, 2,3,4,5,\) or \(6 .\) Use this information to solve each exercise. (image can't copy) What fraction of the outcomes is not less than \(5 ?\)

Will help you prepare for the material covered in the next section. Each exercise involves observing a pattern in the expanded form of the binomial expression \((a+b)^{n}\). $$\begin{array}{l} (a+b)^{1}=a+b \\ (a+b)^{2}=a^{2}+2 a b+b^{2} \\ (a+b)^{3}=a^{3}+3 a^{2} b+3 a b^{2}+b^{3} \\ (a+b)^{4}=a^{4}+4 a^{3} b+6 a^{2} b^{2}+4 a b^{3}+b^{4} \\ (a+b)^{5}=a^{5}+5 a^{4} b+10 a^{3} b^{2}+10 a^{2} b^{3}+5 a b^{4}+b^{5} \end{array}$$ Describe the pattern for the sum of the exponents on the variables in each term.

Exercises \(31-32\) involve a deck of 52 cards. If necessary, refer to the picture of a deck of cards, Figure 10.12 on page 1110 . If you are dealt 3 cards from a shuffled deck of 52 cards, find the probability that all 3 cards are picture cards.

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