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Chapter 12: Problem 75

What is the coefficient of \(x^{n-1}\) in \((x+1)^{n}\) for \(n=2,3\) and \(4 ?\)

Short Answer

Expert verified
Using the binomial theorem, we found that in the expansion of \((x+1)^n\), the coefficient of the \(x^{n-1}\) term is always equal to \(n\). Therefore, for \(n=2,3\), and \(4\), the coefficients of the \(x^{n-1}\) terms are \(2\), \(3\), and \(4\), respectively.

Step by step solution

01

Recall the binomial theorem

The binomial theorem states that for any positive integer \(n\), \((x+y)^n\) can be expanded as follows: \(\displaystyle (x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k\) Where \(\binom{n}{k}\) are the binomial coefficients, also known as "n choose k", and can be calculated using the formula: \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\)
02

Find the coefficient of \(x^{n-1}\) using the binomial theorem

In our case, we have \((x+1)^n\), and we are interested in the term with \(x^{n-1}\), so we want to find the exponent \(k\) where \(n-k = n-1\). Thus, \(k=1\). Now, let's find the binomial coefficient for this term: \(\binom{n}{k} = \binom{n}{1} = \frac{n!}{1!(n-1)!} = n\) So, for \((x+1)^n\), the coefficient of \(x^{n-1}\) is always equal to \(n\).
03

Find the coefficients for the given values of \(n\)

Now, we can substitute the given values of \(n\) and find the coefficients of \(x^{n-1}\) for each case: 1. For \(n=2\): Coefficient of \(x^{2-1} = x^1\) is \(n=2\). 2. For \(n=3\): Coefficient of \(x^{3-1} = x^2\) is \(n=3\). 3. For \(n=4\): Coefficient of \(x^{4-1} = x^3\) is \(n=4\). So, the coefficient of \(x^{n-1}\) in \((x+1)^n\) for \(n=2,3\) and \(4\) is \(2\), \(3\), and \(4\), respectively.

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Key Concepts

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

Binomial Coefficients
To understand the coefficient of a particular term in the expansion of a binomial expression, we rely on binomial coefficients. These coefficients are the numbers that help determine the weight of specific terms in a polynomial expansion. They are often referred to as "n choose k," representing different possible selections.
  • In mathematical terms, these coefficients are expressed as \( \binom{n}{k} \) and are calculated using the formula \( \frac{n!}{k!(n-k)!} \).
  • "n" is the total number of items to choose from, and "k" is the number of items to choose.
Factorial notation, denoted as \( n! \), signifies the product of all positive integers up to n. For instance, \( 3! = 3 \times 2 \times 1 = 6 \).
These binomial coefficients are crucial in expanding binomials, allowing us to determine the contribution of each term depending on where it falls in the expansion.
Polynomial Expansion
The concept of polynomial expansion is essential when dealing with expressions involving powers. The Binomial Theorem provides the framework for expanding expressions like \((x+y)^n\). This expansion reveals a series of terms, each with its own coefficient, which are determined by the binomial coefficients.
When expanding a binomial raised to a power n, the theorem gives:
  • \((x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \)
  • Each term in this series has a binomial coefficient \( \binom{n}{k} \).
Understanding polynomial expansion is key in helping us break down such expressions into manageable solutions, making complex problems simpler to solve. By applying this expansion, we can see how individual terms like \( x^{n-k} \) and \( y^k \) get their respective powers and coefficients.
Coefficient Calculation
Finding specific coefficients within expanded polynomials is a useful skill, particularly in algebra and calculus. In this context, we'll focus on calculating the coefficient of \(x^{n-1}\) in the expression \((x+1)^n\).
Let's apply the binomial theorem to find it:
  • We need the term where \( n-k = n-1 \), indicating that \( k = 1 \).
  • This gives the binomial coefficient \( \binom{n}{1} \), which simplifies to \( n \).
For specific values like \( n=2, 3, 4 \), calculate:
  • When \( n=2 \), the coefficient is \( 2 \).
  • When \( n=3 \), the coefficient is \( 3 \).
  • When \( n=4 \), the coefficient is \( 4 \).
Thus, in this polynomial, the coefficient of \(x^{n-1}\) equals the integer \(n\) itself.

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Most popular questions from this chapter

Refer to the functions \(f(x)\) and \(g(x),\) where the function $$ g(x)=1+\frac{1}{2} x+\frac{3}{8} x^{2}+\frac{5}{16} x^{3} $$ is used to approximate the values of $$ f(x)=\frac{1}{\sqrt{1-x}} $$ Show that \(f(x)\) is undefined at \(x=1\) and \(x=2\), but that \(g(x)\) is defined at these values. Explain why the algebraic operations used to define \(f\) may lead to undefined values, whereas the operations used to define \(g\) will not.

Give polynomials satisfying the given conditions if possible, or say why it is impossible to do so. Two polynomials whose product has degree \(9 .\)

Given that \(\left(3 x^{3}+a\right)\left(2 x^{b}+3\right)=6 x^{7}+c x^{4}+d x^{3}+3\), find possible values for the constants \(a, b, c,\) and \(d\).

Find approximate solutions to $$ 3 x^{3}-2 x^{2}-6 x+4=0 $$ by graphing the polynomial.

Use the identity \((x-1)\left(x^{4}+x^{3}+x^{2}+x+1\right)=x^{5}-1\) to show that \(8^{5}-1\) is divisible by 7 .
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