Problem 65
Question
Find the indicated term of each binomial expansion. Show that \(\left(\begin{array}{l}n \\ n\end{array}\right)=1\) for any positive integer \(n\)
Step-by-Step Solution
Verified Answer
We used the binomial coefficient formula \(\left(\begin{array}{l}n \\\ k\end{array}\right) = \frac{n!}{k!(n-k)!}\) and substituted \(k=n\), giving us \(\left(\begin{array}{l}n \\\ n\end{array}\right) = \frac{n!}{n!(0)!}\). Since \(0!=1\), the expression simplifies to \(\left(\begin{array}{l}n \\\ n\end{array}\right) = \frac{n!}{n!}\), which equals to 1. Hence, \(\left(\begin{array}{l}n \\\ n\end{array}\right)=1\) for any positive integer \(n\).
1Step 1: Binomial Coefficient Formula
The binomial coefficient can be calculated using the following combinatorics formula:
\[\left(\begin{array}{l}n \\\ k\end{array}\right) = \frac{n!}{k!(n-k)!},\]
where \(n!\) denotes the factorial of \(n\) and is calculated as the product of all positive integers up to \(n\). In our case, we need to find the coefficient for \(k = n\), so the expression will be:
\[\left(\begin{array}{l}n \\\ n\end{array}\right) = \frac{n!}{n!(n-n)!}.\]
2Step 2: Substituting k=n
We will now substitute \(k=n\) in our formula and simplify it:
\[\left(\begin{array}{l}n \\\ n\end{array}\right) = \frac{n!}{n!(0)!}.\]
3Step 3: Simplifying the expression
As \(0! = 1\) in combinatorics, we can simplify the expression further:
\[\left(\begin{array}{l}n \\\ n\end{array}\right) = \frac{n!}{n! \times 1}.\]
Now, when dividing \(n!\) by itself, we simply get 1:
\[\left(\begin{array}{l}n \\\ n\end{array}\right) = \frac{n!}{n!} = 1.\]
4Step 4: Conclusion
We have shown that the binomial coefficient \(\left(\begin{array}{l}n \\\ n\end{array}\right)\) is equal to 1 for any positive integer \(n\).
Key Concepts
CombinatoricsFactorialBinomial ExpansionPositive Integers
Combinatorics
Combinatorics is the branch of mathematics that deals with counting, arranging, and combination of objects. It comes in handy when analyzing possibilities in various scenarios. A fundamental part of combinatorics is the concept of a binomial coefficient, often used to calculate the number of ways to choose a set of items from a larger set, regardless of order. This calculation is central to many problems regarding permutations and combinations. By leveraging combinatorics, we can efficiently solve problems involving large datasets or complex structures by breaking them down into manageable parts.
Factorial
The factorial, denoted as \(n!\), represents the product of all positive integers up to a given number \(n\). For instance, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). It's important in various calculations in mathematics, especially in combinatorics.
- If \(n = 0\), then \(0!\) is defined as 1.
Binomial Expansion
Binomial expansion refers to expanding powers of a binomial expression. A binomial expression is simply two terms added together, like \((a+b)\). According to the Binomial Theorem, any power \(n\) of a binomial \((a+b)^n\) can be expanded into a sum involving binomial coefficients:\[(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]Here, \(\binom{n}{k}\) is the binomial coefficient, representing the number of ways the term can be constructed. Binomial expansion is vital in algebra and calculus since it simplifies the multiplication of polynomial expressions and the evaluation of limits, integrals, and other advanced topics.
Positive Integers
Positive integers are the numbers starting from 1 and continuing upwards, such as 1, 2, 3, and so on. They do not include zero, negative numbers, or fractions - only whole numbers greater than zero.
- They form the basis of many mathematical concepts and operations.
- In the context of combinatorics, positive integers often denote the number of items in a set or the specific term being considered.
Other exercises in this chapter
Problem 64
Find the arithmetic mean of each group of numbers. $$38,31,43,40,33$$
View solution Problem 64
Evaluate each sum using a formula for \(S_{n}\). $$\sum_{i=1}^{9}(2 i-14)$$
View solution Problem 65
Use the formula for \(S_{n}\) to find the sum of the terms of each geometric sequence. $$\sum_{i=1}^{4}(-18)\left(-\frac{2}{3}\right)^{i}$$
View solution Problem 65
Find the arithmetic mean of each group of numbers. $$8,7,11,9,12$$
View solution