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Magazine Chapter 6: Problem 6
What is the coefficient of \(x^{7}\) in \((1+x)^{11} ?\)
Short Answer
Expert verified
The coefficient of \(x^7\) is \(330\).
Step by step solution
01
Identify the Binomial Theorem
The Binomial Theorem states that \( (a + b)^n = \begin{pmatrix} n \ k \ \end{pmatrix} a^{n-k} b^k \). Here, we need to find the coefficient of the term containing \(x^7\) in the expansion of \( (1 + x)^{11}\).
02
Set up the general term
For the term involving \(x^k\) in \( (1 + x)^{11}\), the general term is \( \begin{pmatrix} 11 \ k \ \end{pmatrix} x^k \). We are looking for the term where \( k = 7 \), so we have \( \begin{pmatrix} 11 \ 7 \ \end{pmatrix} x^7 \).
03
Calculate the binomial coefficient
The coefficient of \( x^7 \) is given by the binomial coefficient \( \begin{pmatrix} 11 \ 7 \ \end{pmatrix} = \frac{11!}{7!(11-7)!} = \frac{11!}{7!4!} \).
04
Simplify the factorial expression
Simplify the expression \( \frac{11!}{7!4!} \) as follows: \( 11! = 11 \times 10 \times 9 \times 8 \times 7! \), so \( \frac{11!}{7!4!} = \frac{11 \times 10 \times 9 \times 8 \times 7!}{7!4!} \). The \( 7! \) terms cancel out, resulting in \( \frac{11 \times 10 \times 9 \times 8}{4!} \). Calculate \( 4! = 4 \times 3 \times 2 \times 1 = 24 \), giving us \( \frac{11 \times 10 \times 9 \times 8}{24} \).
05
Perform the final multiplication and division
Calculate the result: \( 11 \times 10 = 110 \), \( 110 \times 9 = 990 \), and \( 990 \times 8 = 7920 \). Then, divide \( \frac{7920}{24} = 330 \). Hence, the coefficient of \( x^7 \) is \( 330 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Binomial Coefficient
Let's start with understanding what a binomial coefficient is. The binomial coefficient is denoted as \binom{n}{k} and represents the number of ways to choose k items from n items without regard to the order of selection. It's basically a way to count combinations.
To find the coefficient of a particular term in a binomial expansion like \(1 + x\)^n, we use the binomial coefficient.
The formula for the binomial coefficient is \[\binom{n}{k} = \frac{n!}{k!(n-k)!} \].
The coefficient is found using \( \binom{11}{7} \).
To find the coefficient of a particular term in a binomial expansion like \(1 + x\)^n, we use the binomial coefficient.
The formula for the binomial coefficient is \[\binom{n}{k} = \frac{n!}{k!(n-k)!} \].
- \(n!\) is the factorial of n.
- \(k!\) is the factorial of k.
The coefficient is found using \( \binom{11}{7} \).
Factorials
Understanding factorials is essential for calculating binomial coefficients.
A factorial, denoted as \( n! \), is the product of all positive integers up to n. For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).
In the exercise, we need to calculate \( 11! \), \( 7! \), and \(4!\) to find the binomial coefficient \( \binom{11}{7} \). Using the factorials:
\[\frac{11!}{7!4!} = \frac{11 \times 10 \times 9 \times 8 \times 7!}{7! \times 4!}\]
The \(7!\) terms cancel out, simplifying it to \[ \frac{11 \times 10 \times 9 \times 8}{4!}\] and then,
\(4! = 4 \times 3 \times 2 \times 1 = 24\).
A factorial, denoted as \( n! \), is the product of all positive integers up to n. For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).
In the exercise, we need to calculate \( 11! \), \( 7! \), and \(4!\) to find the binomial coefficient \( \binom{11}{7} \). Using the factorials:
\[\frac{11!}{7!4!} = \frac{11 \times 10 \times 9 \times 8 \times 7!}{7! \times 4!}\]
The \(7!\) terms cancel out, simplifying it to \[ \frac{11 \times 10 \times 9 \times 8}{4!}\] and then,
\(4! = 4 \times 3 \times 2 \times 1 = 24\).
- This simplification makes it easier to perform the multiplication and division to find the final result.
Polynomial Expansion
Polynomial expansion involves expressing a polynomial raised to a power as a sum of terms involving the variables and their powers.
The Binomial Theorem helps us expand \(a + b\)^n into a series of terms. Each term in the expansion is formed using the binomial coefficient and follows the general form:
\[ \binom{n}{k} a^{n-k} b^k \]
In our exercise, \(a = 1 \) and \(b = x, n = 11\). We were looking for the term where the power of x is 7 (\( x^7 \)). The general term we use is:
\( \bimom{11}{7} x^7 \).
The Binomial Theorem helps us expand \(a + b\)^n into a series of terms. Each term in the expansion is formed using the binomial coefficient and follows the general form:
\[ \binom{n}{k} a^{n-k} b^k \]
In our exercise, \(a = 1 \) and \(b = x, n = 11\). We were looking for the term where the power of x is 7 (\( x^7 \)). The general term we use is:
\( \bimom{11}{7} x^7 \).
- Using the binomial coefficients, the polynomial can be expanded in a quick and organized way.
- This is useful especially in combinatorial problems where we need to find specific terms in large expansions.
Combinatorics
Combinatorics is the branch of mathematics dealing with combinations and permutations of objects.
In our exercise about finding the coefficient of a specific term in binomial expansion, combinatorics plays a vital role.
We used combinatorial methods (binomial coefficients) to determine the number of ways to choose 7 items from 11. This directly gave us the coefficient for the term involving \(x^7\).
Combinatorics helps us solve problems related to counting, arranging, and selecting objects in a structured manner.
In our exercise about finding the coefficient of a specific term in binomial expansion, combinatorics plays a vital role.
We used combinatorial methods (binomial coefficients) to determine the number of ways to choose 7 items from 11. This directly gave us the coefficient for the term involving \(x^7\).
Combinatorics helps us solve problems related to counting, arranging, and selecting objects in a structured manner.
- Understanding combinatorics makes it easier to grasp more complex topics in probability, statistics, and algebra.
