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$$ \text { Expand the expressions in 1-9 using the binomial theorem. } $$ $$ (1+x)^{7} $$

Short Answer

Expert verified
Using the binomial theorem, the expansion of \((1+x)^{7}\) is: \((1 + x)^7 = 1 + 7x + 21x^2 + 35x^3 + 35x^4+ 21x^5 + 7x^6 + x^7\)

Step by step solution

01

Recall the binomial theorem

The binomial theorem states that for any non-negative integer n and terms a and b, the expansion of \((a+b)^{n}\) can be given by: \[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\] Here, \(\binom{n}{k}\) represents the binomial coefficient, also known as "n choose k", and can be calculated as: \[\binom{n}{k} = \frac{n!}{k!(n-k)!}\] In our case, we have a = 1, b = x, and n = 7.
02

Determine the terms in the expansion

Using the binomial theorem formula, we can now determine the terms in the expansion of \((1+x)^{7}\): \((1 + x)^7 = \sum_{k=0}^{7} \binom{7}{k} (1)^{7-k} (x)^k\) We will compute each term of the sum as follows: n = 7, k = 0, 1, 2, 3, 4, 5, 6, 7
03

Calculate the binomial coefficients

Compute the binomial coefficients for each term: - \(\binom{7}{0} = \frac{7!}{0!(7-0)!} = 1\) - \(\binom{7}{1} = \frac{7!}{1!(7-1)!} = 7\) - \(\binom{7}{2} = \frac{7!}{2!(7-2)!} = 21\) - \(\binom{7}{3} = \frac{7!}{3!(7-3)!} = 35\) - \(\binom{7}{4} = \frac{7!}{4!(7-4)!} = 35\) - \(\binom{7}{5} = \frac{7!}{5!(7-5)!} = 21\) - \(\binom{7}{6} = \frac{7!}{6!(7-6)!} = 7\) - \(\binom{7}{7} = \frac{7!}{7!(7-7)!} = 1\)
04

Apply the binomial coefficients to find the terms

Now, we will apply the binomial coefficients to find the 8 terms of the expansion: - Term 1: \(\binom{7}{0} (1)^{7-0} (x)^0 = (1)(1)(1) = 1\) - Term 2: \(\binom{7}{1} (1)^{7-1} (x)^1 = (7)(1)(x) = 7x\) - Term 3: \(\binom{7}{2} (1)^{7-2} (x)^2 = (21)(1)(x^2) = 21x^2\) - Term 4: \(\binom{7}{3} (1)^{7-3} (x)^3 = (35)(1)(x^3) = 35x^3\) - Term 5: \(\binom{7}{4} (1)^{7-4} (x)^4 = (35)(1)(x^4) = 35x^4\) - Term 6: \(\binom{7}{5} (1)^{7-5} (x)^5 = (21)(1)(x^5) = 21x^5\) - Term 7: \(\binom{7}{6} (1)^{7-6} (x)^6 = (7)(1)(x^6) = 7x^6\) - Term 8: \(\binom{7}{7} (1)^{7-7} (x)^7 = (1)(1)(x^7) = x^7\)
05

Write the expanded expression

Now that we have all the terms in the expansion, add them together: \((1 + x)^7 = 1 + 7x + 21x^2 + 35x^3 + 35x^4+ 21x^5 + 7x^6 + x^7\)

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Key Concepts

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

Binomial Expansion
Binomial expansion is a method used to express the power of a binomial as a series of terms. In simpler terms, expanding any binomial expression like \((a+b)^n\) into a sum of terms involving powers of \(a\) and \(b\) is called a binomial expansion.
A binomial expression has just two terms, typically written as \((a+b)\).When expanded, each term is a product of the binomial coefficient, a power of \(a\), and a power of \(b\).
The key to expansion is to apply the binomial theorem, which is written as:
  • \( (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \)
In this formula, you've got to calculate \(\binom{n}{k}\), which gives you the coefficients you multiply by each term. These coefficients are essential and are calculated separately.
This makes working with binomial expansions predictable and repeatable, no matter how large the power is.
When you practice this several times, you'll find it becomes systematic and easier to remember! The main takeaway here is that the binomial expansion is simply a way to rewrite a binomial expression raised to a power in a series of terms that are easier to handle.
Binomial Coefficient
A binomial coefficient plays a vital role in binomial expansion. It tells us the number of ways to choose \(k\) elements from a set of \(n\) elements and is often referred to as "n choose k". Mathematically, it's expressed as \(\binom{n}{k}\), which can be calculated using the formula:
  • \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)
The exclamation point (\(!\)) stands for factorial, which is a product of all positive integers up to that number.For example, \(!3! = 3 \times 2 \times 1 = 6\).
This calculation isn't just limited to mathematics; it finds uses in fields like statistics when evaluating probabilities.The coefficients also make Pascal's Triangle, which is a neat way to visualize them.
Pascal’s Triangle can be used to directly read off the coefficients for expansions, providing a helpful mnemonic.What's fascinating is that these coefficients appear in algebra, but also have practical applications in choosing classes or organizing tournaments.
Combinatorics
Combinatorics is an area of mathematics dealing with combinations of objects under specific conditions. It helps in counting, arranging, and understanding the structures within a finite set.It's the mathematics behind creating and manipulating groups of items. The binomial coefficient is a classic example of combinatorics in action - it calculates the combinations of choosing \(k\) items from \(n\) without respecting order.Combinatorics is not just about counting items. It's also about understanding how those items relate under different constraints and conditions.
You see combinatorics at work in probability theory, graph theory, and complex algorithms in computer science. Naturally, it also underpins problems like the one discussed, providing the rationale and calculation method.Being skilled in combinatorics can therefore unlock diverse insights and solutions across various scientific fields. It's essentially the toolkit that allows mathematicians to solve seemingly simple yet intricately complex problems!

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Most popular questions from this chapter

A lottery game offers \(\$ 2\) million to the grand prize winner, \(\$ 20\) to each of 10,000 second prize winners, and \(\$ 4\) to each of 50,000 third prize winners. The cost of the lottery is \(\$ 2\) per ticket. Suppose that \(1.5\) million tickets are sold. What is the expected gain or loss of a ticket?

An urn contains 25 red balls and 15 blue balls. Two are chosen at random, one after the other, without replacement. a. What is the probability that both balls are red? b. What is the probability that the second ball is red but the first ball is not? c. What is the probability that the second ball is red? d. What is the probability that at least one of the balls is red?

An alternative way to derive Theorem 6.4.1 uses the following division rule: Let \(n\) and \(k\) be integers so that \(k\) divides \(n\). If a set consisting of \(n\) elements is divided into subsets that each contain \(k\) elements, then the number of such subsets is \(n / k\). Explain how Theorem \(6.4 .1\) can be derived using the division rule.

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