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Chapter 5: Problem 5

Expand \((2 x-y)^{7}\) using the binomial theorem.

Short Answer

Expert verified
The expanded form of \((2x-y)^{7}\) using the binomial theorem is \((2x)^7 - 7(2x)^6(-y) + 21(2x)^5(-y)^2 - 35(2x)^4(-y)^3 + 35(2x)^3(-y)^4 - 21(2x)^2(-y)^5 + 7(2x)(-y)^6 - (-y)^7\) . After simplifying, this equals \(128x^7 - 448x^6y + 704x^5y^2 - 560x^4y^3 + 280x^3y^4 - 84x^2y^5 + 14xy^6 - y^7\).

Step by step solution

01

Identify the components of the binomial

In the given expression \((2x-y)^{7}\), \(a = 2x\), \(b = -y\), and \(n = 7\). We have to substitute these values into the binomial theorem \((a+b)^{n}= \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}\)
02

Apply the binomial theorem

Find each of the terms in the sum from \(k = 0\) to \(k = 7\). To do this, for each \(k\), calculate the binomial coefficient \(\binom{7}{k}\), multiply by \(2x\) to the power of \(7-k\), and multiply by \(-y\) to the power of \(k\). Ensuring that you follow the rules of multiplying with exponents when calculating the powers of \(2x\) and \(-y\), and remember that any number to the power of 0 is 1.
03

Simplify each term and write out the expansion

Combine the calculated values to write out the full expansion of \((2x-y)^{7}\). Each term will look like \(\binom{7}{k} (2x)^{7-k} (-y)^{k}\) for a given \(k\), and the entire expression will be a sum of such terms for \(k = 0\) to \(k = 7\).

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Key Concepts

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

Binomial Coefficients
Binomial coefficients are a key part of the Binomial Theorem. They are represented using the symbol \( \binom{n}{k} \) and are read as "n choose k." Binomial coefficients count the number of ways to choose \(k\) elements from a set of \(n\) distinct elements.
  • When you see a binomial expression like \((a + b)^n\), binomial coefficients help determine the coefficient for each term in the expanded expression.
  • You can calculate a binomial coefficient using the formula \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\), where \(!\) denotes a factorial, which is the product of an integer and all the integers below it.
  • In the expression \((2x-y)^7\), you will calculate the binomial coefficient for each term, starting from \(k = 0\) to \(k = 7\).
These coefficients are important because they guide the multiplication of the terms \(a\) and \(b\) raised to their respective powers in each stage of the expansion.
Polynomial Expansion
Polynomial expansion using the binomial theorem involves expressing a power of a binomial as a sum of terms. Each term in the expansion of \((a + b)^n\) consists of:
  • A binomial coefficient \(\binom{n}{k}\)
  • The term \(a\) raised to the power of \(n-k\)
  • The term \(b\) raised to the power of \(k\)
In the problem \((2x-y)^7\), the binomial expansion will lead to a polynomial with several terms. Each term is uniquely constructed by varying \(k\) from \(0\) to \(n\), giving each term its own combination of coefficients and powers. The process is systematic, ensuring that every possible combination of powers is covered, leading to a complete expansion of the original binomial.
Exponentiation
Exponentiation refers to the operation of raising a number to a power. In the context of binomial expansion, exponentiation is present in both the terms being expanded and the resulting expanded terms themselves.
  • For example, in \((2x-y)^7\), exponentiation starts with the whole expression being raised to the seventh power.
  • Click by click, small pieces or terms like \( (2x)^{7-k} \) and \( (-y)^k \) are created. Each term requires careful computation because powers must be respected for both positive and negative numbers.
  • Remember that any number raised to the power of 0 is 1, which simplifies calculations when determining terms.
Exponentiation is a basic mathematical operation, but its correct application in binomial expansions ensures that answers are accurate and consistent.
Combinatorial Mathematics
Combinatorial mathematics is deeply embedded in the binomial theorem, especially when calculating the binomial coefficients.
  • The theorem is a beautiful mix of applying algebra to solve combinatorial counting problems through coefficients \( \binom{n}{k} \), which are essential to expanding the polymer correctly.
  • Each term \( \binom{n}{k} (a^{n-k} b^k) \) results from selecting, combining, and arranging pieces of expressions, much like how you'd arrange objects in different sequences or groups.
  • Combinatorics offers several helpful concepts, like permutations and combinations, which aid in the understanding of selection and arrangement principles.
Overall, combinatorial mathematics provides the structural underpinning for calculating and using binomial coefficients, which in turn makes polynomial expansion feasible.

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Most popular questions from this chapter

Expand \(\left(x_{1}+x_{2}+x_{3}\right)^{n}\) by observing that $$ \left(x_{1}+x_{2}+x_{3}\right)^{n}=\left(\left(x_{1}+x_{2}\right)+x_{3}\right)^{n} $$ and then using the binomial theorem.

Prove that $$ \sum_{n_{1}+n_{2}+n_{3}=n}\left(\begin{array}{cc} n & \\ n_{1} n_{2} n_{3} \end{array}\right)(-1)^{n_{1}-n_{2}+n_{3}}=1 $$ where the summation extends over all nonnegative integral solutions of \(n_{1}+n_{2}+\) \(n_{3}=n .\)

What is the coefficient of \(x_{1}^{3} x_{2}^{3} x_{3} x_{4}^{2}\) in the expansion of $$ \left(x_{1}-x_{2}+2 x_{3}-2 x_{4}\right)^{9} ? $$

Evaluate the sum $$ \sum_{k=0}^{n}(-1)^{k}\left(\begin{array}{l} n \\ k \end{array}\right) 10^{k} . $$

Let \(R\) and \(S\) be two partial orders on the same set \(X\). Considering \(R\) and \(S\) as subsets of \(X \times X\), we assume that \(R \subseteq S\) but \(R \neq S\). Show that there exists an ordered pair \((p, q)\), where \((p, q) \in S\) and \((p, q) \notin R\) such that \(R^{\prime}=R \cup\\{(p, q)\\}\) is also a partial order on \(X\). Show by example that not every such \((p, q)\) has the property that \(R^{\prime}\) is a partial order on \(X\).
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