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Without expanding completely, find the indicated term(s) in the expansion of the expression. \(\left(\frac{3}{c}+\frac{c^{2}}{4}\right)^{7} ; \quad\) sixth term

Short Answer

Expert verified
The sixth term is \(\frac{189 c^8}{1024}\).

Step by step solution

01

Understand the Binomial Theorem

The Binomial Theorem states that \((a + b)^n\) can be expanded as \(\sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). Here, the expression is \(\left(\frac{3}{c}+\frac{c^2}{4}\right)^7\). We are looking for the sixth term in the expansion.
02

Identify the Elements for the Binomial Term Formula

In our expression, \(a = \frac{3}{c}\) and \(b = \frac{c^2}{4}\), with \(n = 7\). The sixth term corresponds to \(k = 5\) (since the first term is for \(k = 0\)).
03

Apply the Term Formula

The sixth term is given by \(T_{k+1} = \binom{n}{k} a^{n-k} b^k\). Substitute \(n = 7\), \(k = 5\), \(a = \frac{3}{c}\), and \(b = \frac{c^2}{4}\) into the formula.
04

Calculate the Binomial Coefficient

The binomial coefficient is \(\binom{7}{5}\), which equals \(21\).
05

Calculate Powers of a and b

Calculate \(a^{n-k} = \left(\frac{3}{c}\right)^{7-5} = \left(\frac{3}{c}\right)^2 = \frac{9}{c^2}\) and \(b^k = \left(\frac{c^2}{4}\right)^5 = \frac{c^{10}}{1024}\).
06

Combine All Parts to Get the Term

Combine the binomial coefficient and the powers together: \(T_6 = 21 \times \frac{9}{c^2} \times \frac{c^{10}}{1024} = \frac{21 \times 9 \times c^{8}}{1024}\).
07

Simplify the Expression

Calculate and simplify the coefficients: \(21 \times 9 = 189\). Thus, the sixth term is \(\frac{189 c^8}{1024}\).

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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 vital part of the Binomial Theorem, which allows us to expand expressions of the form \((a + b)^n\). It is usually denoted as \(\binom{n}{k}\), pronounced "n choose k". This coefficient is simply a way to describe the number of different ways you can choose \(k\) items out of a total of \(n\). To calculate this, we use the formula:
  • \(\binom{n}{k} = \frac{n!}{k!\,(n-k)!}\)
where \(!\) denotes factorial, which is the product of all positive integers less than or equal to a number. When dealing with expressions like the one in our original problem, the Binomial Coefficient helps us to find specific terms in the expanded form without having to expand the entire expression. This is extremely useful for simplifying and solving polynomial expressions.
Polynomial Expansion
Polynomial expansion is the process of expressing a polynomial raised to a power, like \((a + b)^7\), in a sum of terms. Each term is a product of a binomial coefficient and powers of both \(a\) and \(b\). For example, each term in the expansion of \((a + b)^n\) is given by
  • \(T_{k+1} = \binom{n}{k} a^{n-k} b^k\)
By following this formula, we can accurately determine each separate term in the expanded polynomial. The benefit of polynomial expansion in algebra is that it simplifies the process of working with power expressions by breaking them down into manageable parts. It showcases the powerful application of the Binomial Theorem to higher degree polynomials.Understanding polynomial expansion is fundamental in algebra. It opens the door to solving more complex algebraic problems and provides a clear method to simplify expressions.
Algebraic Expressions
Algebraic expressions are an essential part of mathematics, encompassing combinations of numbers, variables, and operators such as addition, subtraction, and multiplication. They form the building blocks for more complex equations and serve as a language through which mathematicians and scientists describe quantities and their relationships. In our example, the expression \(\left(\frac{3}{c}+\frac{c^{2}}{4}\right)^{7}\) is an algebraic expression involving both fractions and variables raised to a power. Here's why understanding algebraic expressions is important:
  • They allow us to generalize concepts, making it possible to work with unknowns and find solutions for variables.
  • Through manipulation of these expressions, we can simplify and solve equations, model real-world scenarios, and even predict outcomes in various scientific fields.
These expressions form a critical component of many mathematical operations, from simple solving to complex engineering problems, highlighting their universal applicability.

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Most popular questions from this chapter

Card and die experiment Each suit in a deck is made up of an ace (A), nine numbered cards \((2,3, \ldots, 10),\) and three face cards (J, Q, K). An experiment consists of drawing a single card from a deck followed by rolling a single die. Describe the sample space \(S\) of the experiment, and find \(n(S)\) Let \(E_{1}\) be the event consisting of the outcomes in which a numbered card is drawn and the number of dots on the die is the same as the number on the card. Find \(n\left(E_{1}\right), n\left(E_{1}^{\prime}\right),\) and \(P\left(E_{1}\right)\) Let \(E_{2}\) be the event in which the card drawn is a face card, and let \(E_{3}\) be the event in which the number of dots on the die is even. Are \(E_{2}\) and \(E_{3}\) mutually exclusive? Are they independent? Find \(P\left(E_{2}\right), P\left(E_{3}\right)\) \(P\left(E_{2} \cap E_{3}\right),\) and \(P\left(E_{2} \cup E_{3}\right)\) Are \(E_{1}\) and \(E_{2}\) mutually exclusive? Are they independent? Find \(P\left(E_{1} \cap E_{2}\right)\) and \(P\left(E_{1} \cup\right.\) \(\left.E_{2}\right)\)

Depreciation methods are sometimes used by businesses and individuals to estimate the value of an asset over a life span of \(n\) years. In the sum-of- year's-digits method, for each year \(k=1,2,3, \ldots, n,\) the value of an asset is decreased by the fraction \(A_{k}=\frac{n-k+1}{T_{n}}\) of its initial cost, where \(T_{n}=1+2+3+\cdots+n\). (a) If \(n=8,\) find \(A_{1}, A_{2}, A_{3}, \dots, A_{3}\) (b) Show that the sequence in (a) is arithmetic, and find \(S_{\mathrm{g}}\). (c)If the initial value of an asset is \(\$ 1000,\) how much has been depreciated after 4 years?

Work Exercise 35 if the digits are \(2,4,\) and 7 and one of these digits is repeated in the four-digit code.

The equation \(\frac{1}{3} \sqrt[3]{x}-x+2=0\) has a root near \(2 .\) To approximate this root, rewrite the equation as \(x=\frac{1}{3} \sqrt[3]{x}+2\) Let \(x_{1}=2\) and find successive approximations \(x_{2}, x_{3}, \ldots\) by using the formulas $$ x_{2}=\frac{1}{3} \sqrt[3]{x_{1}}+2, \quad x_{3}=\frac{1}{3} \sqrt[3]{x_{2}}+2, \quad \ldots $$ until four-decimal-place accuracy is obtained.

Chlorine tevels Chlorine is often added to swimming pools to control microorganisms. If the level of chlorine rises above 3 ppm (parts per million), swimmers will experience burning eyes and skin discomfort. If the level drops below 1 ppm, there is a possibility that the water will turn green because of a large algae count. Chlorine must be added to pool water at regular intervals. If no chlorine is added to a pool during a 24 -hour period, approximately \(20 \%\) of the chlorine will dissipate into the atmosphere and \(80 \%\) will remain in the water. (a) Determine a recursive sequence \(a_{n}\) that expresses the amount of chlorine present after \(n\) days if the pool has \(a_{0}\) ppm of chlorine initially and no chlorine is added. (b) If a pool has 7 ppm of chlorine initially, construct a table to determine the first day on which the chlorine level will drop below 3 ppm.

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