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Chapter 10: Problem 48

Find the indicated term in the expansion of the given expression. Third term of \((x-5)^{5}\)

Short Answer

Expert verified
The third term of the expansion is \(250x^3\).

Step by step solution

01

Understand the Formula

To find a specific term in an expansion, we use the Binomial Theorem, which states: \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). Here, \(a = x\), \(b = -5\), and \(n = 5\).
02

Determine the Term Number

Since we are interested in the third term, we use the expression \(k+1 = 3\) to find \(k\). Solving for \(k\), we have \(k = 2\).
03

Apply the Binomial Coefficient

Use \(k = 2\) in the Binomial Theorem to find the third term: \(\binom{5}{2} x^{5-2} (-5)^2\).
04

Calculate the Binomial Coefficient

Calculate \(\binom{5}{2}\), which equals \(\frac{5 \times 4}{2 \times 1} = 10\).
05

Compute the Exponent Values

Calculate \(x^{5-2} = x^3\) and \((-5)^2 = 25\).
06

Put It All Together

Multiply all parts of the third term: \(10 \times x^3 \times 25 = 250x^3\).

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Key Concepts

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

Binomial Coefficient
The binomial coefficient is a fundamental part of the Binomial Theorem. It is represented as \( \binom{n}{k} \), which is called "n choose k" and is used to determine the coefficient of a specific term in a binomial expansion. The formula for the binomial coefficient is \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \( n! \) (n factorial) is the product of all positive integers from 1 to \( n \). This concept is crucial for finding specific terms in the expansion of expressions like \((x - 5)^5\).
  • It determines how many ways you can choose \( k \) elements from a total of \( n \) elements.
  • The binomial coefficient helps in calculating the value that multiplies the variables in each term of the polynomial expansion.
  • For example, in the binomial expansion of \((x-5)^5\), if you need the third term, you calculate the coefficient using \( \binom{5}{2} \).
The calculated value gives the coefficient which is then used to find the term \( 10x^3(-5)^2 \), leading to the full third term in the polynomial expansion.
Polynomial Expansion
Polynomial expansion refers to the process of expanding expressions that are raised to a power, such as \((x - 5)^5\). Using the Binomial Theorem, we expand such polynomials into a sum of terms, making it easier to solve or simplify the expression. For the expansion of a binomial \((a + b)^n\), every term in the expansion has a distinct coefficient, power of \( a \), and power of \( b \).
  • The coefficients are determined using binomial coefficients.
  • Each term in the expanded form can be expressed as \( \binom{n}{k} a^{n-k} b^k \).
  • In \((x - 5)^5\), if we build the expression term by term, we start from \( (x)^5 \) down to \( (-5)^5 \).
This expansion is important in algebra as it allows mathematicians to handle complex polynomial expressions in simpler, manageable terms by addressing smaller components.
Exponentiation
Exponentiation is the mathematical operation involving a base and an exponent or power. In the context of binomial expansions, exponentiation is used to express powers of different elements, such as \( x \) and constants like \( -5 \). The expanded polynomial terms like \( x^{5-2} \) and \((-5)^2\) are examples of exponentiation.
  • The base is the number or variable that is repeatedly multiplied.
  • The exponent indicates how many times the base is multiplied by itself.
  • For instance, in \( x^3 \), \( x \) is multiplied three times: \( x \times x \times x \).
  • Similarly, \((-5)^2\) means \(-5 \times -5\), resulting in 25.
Understanding exponentiation is vital because it allows the accurate calculation of each term when expanding binomials. It simplifies calculations, leading to correct and coherent evaluations of terms in polynomial expansions.

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Most popular questions from this chapter

The given recursively-defined sequence \(\left\\{a_{n}\right\\}\) is known to converge. If \(L\) denotes the limit of the sequence, then we must have \(\lim _{n \rightarrow \infty} a_{n}=L\) and \(\lim _{n \rightarrow \infty} a_{n+1}=L\). Use these facts and the recursion formula to find the value of \(L\). $$ a_{1}=1, a_{n+1}=1+\frac{1}{1+a_{n}} $$

Evaluate \(P(n, r)\). $$ P(6,4) $$

Determine whether the given sequence converges. $$ \left\\{\frac{1}{5 n+6}\right\\} $$

The infinite series \(\sum_{k=1}^{\infty} \frac{1}{k(k+1)}\) is an example of a telescoping series. For such series it is possible to find a formula for the general term \(S_{n}\) of the sequence of partial sums. (a) Use the partial fraction decomposition $$ \frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1} $$ as an aid in finding a formula for \(S_{n}\). This will also explain the meaning of the word telescoping. (b) Use part (a) to find the sum of the infinite series.

Evaluate \(C(n, r)\). $$ C(13,11) $$
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