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$$ \text { Expand the expressions in 1-9 using the binomial theorem. } $$ $$ (u-v)^{5} $$

Short Answer

Expert verified
The short answer for the expansion of \((u-v)^5\) using the binomial theorem is: \((u-v)^5 = u^5 - 5u^4v + 10u^3v^2 - 10u^2v^3 + 5uv^4 - v^5\)

Step by step solution

01

Identify a, b, and n in the given expression

In our given expression, \((u-v)^5\), we have: - a = u - b = -v - n = 5 Now we will substitute these values into the binomial theorem formula.
02

Write the formula for binomial expansion

Using the binomial theorem formula, the expansion of \((u-v)^5\) can be expressed as: \((u-v)^5 = \sum_{k=0}^{5} \binom{5}{k} u^{5-k} (-v)^{k}\)
03

Expand each term in the summation

Now we will expand the summation by calculating each term for k = 0, 1, 2, 3, 4, 5. When k = 0: \(\binom{5}{0} u^{5-0} (-v)^{0} = 1 * u^5 * 1 = u^5\) When k = 1: \(\binom{5}{1} u^{5-1} (-v)^{1} = 5 * u^4 * (-v) = -5u^4v\) When k = 2: \(\binom{5}{2} u^{5-2} (-v)^{2} = 10 * u^3 * v^2 = 10u^3v^2\) When k = 3: \(\binom{5}{3} u^{5-3} (-v)^{3} = 10 * u^2 * (-v^3) = -10u^2v^3\) When k = 4: \(\binom{5}{4} u^{5-4} (-v)^{4} = 5 * u * v^4 = 5uv^4\) When k = 5: \(\binom{5}{5} u^{5-5} (-v)^{5} = 1 * 1 * (-v^5) = -v^5\)
04

Combine terms to find the expansion

Now we will combine the terms we calculated to find the expanded expression of \((u-v)^5\). \((u-v)^5 = u^5 - 5u^4v + 10u^3v^2 - 10u^2v^3 + 5uv^4 - v^5\) The expansion of the given expression, \((u-v)^5\), using the binomial theorem is: \(u^5 - 5u^4v + 10u^3v^2 - 10u^2v^3 + 5uv^4 - v^5\)

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Key Concepts

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

Binomial Coefficients
Understanding binomial coefficients is key to demystifying the binomial theorem expansion. A binomial coefficient, symbolized by the notation \(\binom{n}{k}\), represents the number of ways you can choose k items from a set of n distinct items. These coefficients are central to the binomial theorem as they determine the weights of the terms in the expansion of a binomial expression like \( (u-v)^5 \).

When we expand this expression using the binomial theorem, we calculate the binomial coefficients for each term. For example, \(\binom{5}{0}\) equals 1, as there is one way to select zero items from five. Similarly, \(\binom{5}{1}\) equals 5, meaning there are five ways to choose one item from five. These coefficients follow a pattern known as Pascal's Triangle.

Here's a simplified breakdown for visual learners:
  • \binom{5}{0}\ (top of the triangle) - always 1
  • \binom{5}{1}\ - corresponds to the first line of the triangle
  • ...up to
  • \binom{5}{5}\ (bottom of the triangle) - also always 1
The symmetrical nature of Pascal's Triangle means coefficients are mirrored along its center, for instance, \(\binom{n}{k} = \binom{n}{n-k}\). This is why in expansion we see \(\binom{5}{2}\) and \(\binom{5}{3}\) both equal 10.
Polynomial Expression
A polynomial expression consists of variables, coefficients, and exponents combined using addition, subtraction, multiplication, and non-negative integer exponents of variables. In the context of the binomial theorem, the expansion of a binomial, such as \( (u-v)^5 \), results in a polynomial.

Breaking down the terms of our example, \( u^5 - 5u^4v + 10u^3v^2 - 10u^2v^3 + 5uv^4 - v^5 \), each separate expression like \( 5u^4v \) is a term of the polynomial. The highest exponent in the polynomial indicates its degree; thus \( (u-v)^5 \) is a fifth-degree polynomial. These polynomials not only have numerical coefficients but also follow a pattern where the power of u begins at n (our original exponent) and decreases to zero, while the power of v increases from zero to n.

Part of the beauty of polynomial expressions is their predictability. The expanded form of a binomial raised to any integer power can be quickly constructed once you know the binomial coefficients, allowing for a consistent and organized approach to solving polynomial equations.
Summation Notation
Summation notation—indicated by the Greek letter Sigma (Σ)—is a convenient way to express the addition of a sequence of values. It is often used in the binomial theorem to simplify the representation of polynomial expansion. This compact form captures a potentially lengthy process in a single expression.

For instance, the summation notation for our binomial expression \( (u-v)^5 \) is written as \[\sum_{k=0}^{5} \binom{5}{k} u^{5-k} (-v)^{k}\]. This tells us to sum the terms when k takes on each integer value from 0 to 5. Each term consists of a binomial coefficient \(\binom{5}{k}\), a power of u (decreasing as k increases), and a power of v (increasing as k increases).

When using summation notation, it’s vital to carefully calculate each term for every value of k within the stated range, following the binomial theorem pattern.

Application of Summation in Binomial Expansion

Going through the summation step by step, we identify each term, ensuring we consider the alternating signs by including \( (-v)^k \). Combining these calculated terms yields the full expansion of the binomial expression. Summation notation is immensely useful for students and mathematicians alike as it simplifies complex polynomial expansions into a manageable and systematic process.

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Most popular questions from this chapter

A large pile of coins consists of pennies, nickels, dimes, and quarters (at least 30 of each). a. How many different collections of 30 coins can be chosen? b. What is the probability that a collection of 30 coins chosen at random will contain at least four coins of each type?

a. A bit string is a finite sequence of 0 's and 1 's. How many bit strings have length 8 ? b. How many bit strings of length 8 begin with three 0 's? c. How many bit strings of length 8 begin and end with a 1 ? d. In Section \(1.5\) we showed how integers can be represented by strings of 0 's and 1 's inside a digital computer. In fact, through various coding schemes, strings of 0 's and l's can be used to represent all kinds of symbols. One commonly used code is the Extended Binary-Coded Decimal Interchange Code (EBCDIC) in which each symbol has an 8-bit representation. How many distinct symbols can be represented by this code?

A coin is loaded so that the probability of heads is \(0.7\) and the probability of tails is \(0.3\). Suppose that the coin is tossed twice and that the results of the tosses are independent. a. What is the probability of obtaining exactly two heads? b. What is the probability of obtaining exactly one head? c. What is the probability of obtaining no heads? d. What is the probability of obtaining at least one head?

Two faces of a six-sided die are painted red, two are painted blue, and two are painted yellow. The die is rolled three times, and the colors that appear face up on the first, second, and third rolls are recorded. a. Let \(B B R\) denote the outcome where the color appearing face up on the first and second rolls is blue and the color appearing face up on the third roll is red. Because there are as many faces of one color as of any other, the outcomes of this experiment are equally likely. List all 27 possible outcomes. b. Consider the event that all three rolls produce different colors. One outcome in this event is \(R B Y\) and another \(R Y B\). List all outcomes in the event. What is the probability of the event? c. Consider the event that two of the colors that appear face up are the same. One outcome in this event is \(R R B\) and another is \(R B R\). List all outcomes in the event. What is the probability of the event?

a. How many distinguishable ways can the letters of the word \(M I L L I M I C R O N\) be arranged? b. How many distinguishable arrangements of the letters of \(M I L L I M I C R O N\) begin with \(M\) and end with \(N\) ? c. How many distinguishable arrangements of the letters of MILLIMICRON contain the letters \(C R\) next to each other in order and also the letters \(O N\) next to each other in order?

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