Problem 25
Question
In Exercises 19 - 40, use the Binomial Theorem to expand and simplify the expression. \( \left(x + y\right)^5 \)
Step-by-Step Solution
Verified Answer
The simplified form of \((x + y)^5\) is \(x^5 + 5x^{4}y + 10x^{3}y^2 + 10x^{2}y^3 + 5xy^{4} + y^5\).
1Step 1: Apply the Binomial Theorem
First apply the Binomial Theorem to expand \((x + y)^5\). The nth term in the binomial expansion will be \(\binom{5}{k} x^{5-k} y^{k}\).
2Step 2: Write out each term
The expansion of \((x + y)^5\) will be \(x^5 + \binom{5}{1}x^{4}y + \binom{5}{2}x^{3}y^2 + \binom{5}{3}x^{2}y^3 + \binom{5}{4}xy^{4} + y^5\)
3Step 3: Evaluate the binomial coefficients
Evaluate the binomial coefficients. \( \binom{5}{1} = 5\), \( \binom{5}{2} = 10\), \( \binom{5}{3} = 10\), \( \binom{5}{4} = 5\).
4Step 4: Simplify the expression
After simplifying, you get \(x^5 + 5x^{4}y + 10x^{3}y^2 + 10x^{2}y^3 + 5xy^{4} + y^5\).
Key Concepts
Understanding Binomial CoefficientsExploring Polynomial ExpansionCombinatorics Connection
Understanding Binomial Coefficients
Binomial coefficients are central in calculating and expanding expressions using the Binomial Theorem. They are represented by the notation \( \binom{n}{k} \), which means "n choose k". This represents the number of ways to choose \(k\) elements from a set of \(n\) elements.
Mathematically, binomial coefficients are calculated as:\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
where \(n!\) ("n factorial") is the product of all positive integers up to \(n\). These coefficients play a crucial role in determining the weight of each term in a binomial expansion.
For example, in the expansion of \((x + y)^5\), the coefficients \(\binom{5}{1} = 5\), \(\binom{5}{2} = 10\), etc., determine how each term builds up to the complete expanded polynomial.
Mathematically, binomial coefficients are calculated as:\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
where \(n!\) ("n factorial") is the product of all positive integers up to \(n\). These coefficients play a crucial role in determining the weight of each term in a binomial expansion.
For example, in the expansion of \((x + y)^5\), the coefficients \(\binom{5}{1} = 5\), \(\binom{5}{2} = 10\), etc., determine how each term builds up to the complete expanded polynomial.
Exploring Polynomial Expansion
Polynomial expansion transforms a binomial expression raised to a power into a sum of terms. Through the Binomial Theorem, \((x + y)^n\) can be expanded into \(n+1\) terms, each composed of combinations of powers of \(x\) and \(y\).
Each term in the expansion takes the form \(\binom{n}{k}x^{n-k}y^k\), allowing us to see all the ways that two variables can be combined to form terms of degree \(n\). For \((x + y)^5\), when expanded, it becomes:
Each term in the expansion takes the form \(\binom{n}{k}x^{n-k}y^k\), allowing us to see all the ways that two variables can be combined to form terms of degree \(n\). For \((x + y)^5\), when expanded, it becomes:
- \(x^5\), where \(x\) holds the full power
- \(5x^4y\), showing a decrease in power of \(x\) and an increase in power of \(y\)
- \(10x^3y^2\), indicating further progression
Combinatorics Connection
Combinatorics is a field of mathematics focused on counting combinations and arrangements. It forms the backbone of understanding how binomial coefficients are utilized within the Binomial Theorem.
The concept of "combinations" plays a significant role here. When expanding a binomial expression like \((x+y)^n\), the coefficients arise from the different ways we can arrange the powers of \(x\) and \(y\), counted by \(\binom{n}{k}\). Here, each combination represents a distinct term in the expansion.
This deep connection between combinatorics and the Binomial Theorem allows us to systematically and efficiently expand binomials without directly multiplying out everything, saving both time and effort, especially for large powers.
The concept of "combinations" plays a significant role here. When expanding a binomial expression like \((x+y)^n\), the coefficients arise from the different ways we can arrange the powers of \(x\) and \(y\), counted by \(\binom{n}{k}\). Here, each combination represents a distinct term in the expansion.
This deep connection between combinatorics and the Binomial Theorem allows us to systematically and efficiently expand binomials without directly multiplying out everything, saving both time and effort, especially for large powers.
Other exercises in this chapter
Problem 25
In Exercises 25 - 30, find the probability for the experiment of tossing a six-sided die twice. The sum is \( 6 \).
View solution Problem 25
A combination lock will open when the right choice of three numbers (from \( 1 \) to \( 40 \), inclusive) is selected. How many different lock combinations are
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In Exercises 25 - 30, prove the inequality for the indicated integer values of \( n \). \( n! > 2^n, n \ge 4 \)
View solution Problem 25
In Exercises 17 - 28, write the first five terms of the geometric sequence \( a_1 = 3, r = \sqrt{5} \)
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