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Chapter 5: Problem 4

Expand \((x+y)^{5}\) and \((x+y)^{6}\) using the binomial theorem.

Short Answer

Expert verified
(x+y)^5 = x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5, (x+y)^6 = x^6 + 6x^5y + 15x^4y^2 + 20x^3y^3 + 15x^2y^4 + 6xy^5 + y^6

Step by step solution

01

Understanding the binomial theorem

Recall that the binomial theorem allows us to expand power of a binomial as follows: (a+b)^n = 危((n choose k) * a^(n-k) * b^k), where the sum is from k=0 to n. The term (n choose k) = n!/((n-k)! * k!), where '!' denotes the factorial operation.
02

Expand (x+y)^5 using binomial theorem

Substitute a=x, b=y, and n=5 into the binomial theorem. Hence, 危((5 choose k) * x^(5-k) * y^k), where the sum is from k=0 to 5, equals x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5.
03

Expand (x+y)^6 using binomial theorem

Substitute a=x, b=y, and n=6 into the binomial theorem. Hence, 危((6 choose k) * x^(6-k) * y^k), where the sum is from k=0 to 6, equals x^6 + 6x^5y + 15x^4y^2 + 20x^3y^3 + 15x^2y^4 + 6xy^5 + y^6.

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Key Concepts

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

Binomial Expansion
The binomial expansion is a technique that helps us expand expressions like \((a + b)^n\).
This is crucial when dealing with powers of binomials, allowing us to write out the full expanded form of these expressions. The formula for binomial expansion is given by the binomial theorem:
  • \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\)
The symbol \(\binom{n}{k}\) is known as the combinatorial coefficient, which helps in distributing the terms evenly.
This method can be applied to various exponents, making it easier to calculate terms for larger powers.
Factorials in Combinatorics
Factorials play a significant role in combinatorics, especially when calculating combinatorial coefficients.
The factorial of a number \(n\), denoted as \(n!\), is the product of all positive integers up to \(n\). For instance, \(4! = 4 \times 3 \times 2 \times 1 = 24\).
Factorials are used extensively to calculate \(\binom{n}{k}\), defined as \(\frac{n!}{(n-k)!k!}\). This formula is vital for determining the number of ways to choose \(k\) elements from \(n\) elements without regard to order. By understanding factorials, you can simplify complex expressions and solve various combinatorial problems efficiently.
Pascal's Triangle
Pascal's Triangle is a useful tool when dealing with binomial expansions. It visually represents the coefficients of the binomial expansion, making calculations more accessible.
Each row in Pascal's Triangle corresponds to the powers of \(n\), with each number in a row representing the coefficients for corresponding terms in the expansion. For example, the fifth row of the triangle is \(1, 5, 10, 10, 5, 1\), which matches the coefficients in the expansion of \((x + y)^5\).
  • The triangle starts with a single \(1\) at the top.
  • Each subsequent row is constructed by adding the numbers from the row directly above.
This simple yet powerful tool helps in determining the coefficients quickly, without manual computation.
Combinatorial Coefficients
Combinatorial coefficients, often represented as \(\binom{n}{k}\), are used to identify the number of ways to choose \(k\) elements from \(n\) elements without considering the order.
These coefficients are integral in the binomial theorem, providing the weights for each term in the expansion. They follow the identity:
  • \(\binom{n}{k} = \frac{n!}{(n-k)!k!}\)
This tells how many combinations are possible for given \(n\) and \(k\).
The symmetry property \(\binom{n}{k} = \binom{n}{n-k}\) helps in reducing calculations when dealing with binomial expansions. Understanding these coefficients will make solving problems involving combinations or expansions much more manageable.

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Most popular questions from this chapter

What is the coefficient of \(x^{5} y^{13}\) in the expansion of \((3 x-2 y)^{18} ?\) What is the coefficient of \(x^{8} y^{9} ?\) (There is not a misprint in this last question!)

Use the multinomial theorem to show that, for positive integers \(n\) and \(t\). $$ t^{n}=\sum\left(\begin{array}{c} n \\ n_{1} n_{2} \cdots n_{t} \end{array}\right) $$, where the summation extends over all nonnegative integral solutions \(n_{1}, n_{2}, \ldots, n_{t}\) of \(n_{1}+n_{2}+\cdots+n_{t}=n\).

Consider the partially ordered set \((X, \mid)\) on the set \(X=\\{1,2, \ldots, 12\\}\) of the first 12 positive integers, partially ordered by "is divisible by." (a) Determine a chain of largest size and a partition of \(X\) into the smallest. number of antichains. (b) Determine an antichain of largest size and a partition of \(X\) into the smallest number of chains.

\- Let \(S\) be a set of \(n\) elements. Prove that, if \(n\) is even, the only antichain of size \(\left(\begin{array}{l}\mathrm{n} \\\ \left.\frac{n}{2}\right]\end{array}\right)\) is the antichain of all \(\frac{n}{2}\) -subsets; if \(n\) is odd, prove that the only antichains of this size are the antichain of all \(\frac{n-1}{2}\) -subsets and the antichain of all \(\frac{n+1}{2}\) -subsets.

Prove by induction on \(n\) that, for \(n\) a positive integer, $$ \frac{1}{(1-z)^{n}}=\sum_{k=0}^{\infty}\left(\begin{array}{c} n+k-1 \\ k \end{array}\right) z^{k}, \quad|z|<1 . $$ Assume the validity of $$ \frac{1}{1-z}=\sum_{k=0}^{\infty} z^{k}, \quad|z|<1 . $$
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