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Chapter 6: Problem 5

Use the binomial theorem to find the expansion of \((3-2 x)^{6}\) up to and including the term in \(x^{3}\).

Short Answer

Expert verified
The expansion is \(729 - 1458x + 3240x^2 - 4320x^3\) up to \(x^3\).

Step by step solution

01

Understand the Binomial Theorem

The binomial theorem states that \[(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\] where \(\binom{n}{k}\) is a binomial coefficient. For this problem, we will identify \(a = 3\), \(b = -2x\), and \(n = 6\).
02

Apply the Binomial Theorem

Using the binomial theorem, substitute \(a = 3\), \(b = -2x\), and \(n = 6\) into the formula. We need to find terms up to \(x^3\).\[(3-2x)^6 = \sum_{k=0}^{6} \binom{6}{k} (3)^{6-k} (-2x)^k\].
03

Calculate Binomial Coefficients and Terms

We will calculate each term up to \(k=3\):- For \(k=0\): \( \binom{6}{0} (3)^6 (-2x)^0 = 729\)- For \(k=1\): \( \binom{6}{1} (3)^5 (-2x)^1 = -1458x\)- For \(k=2\): \( \binom{6}{2} (3)^4 (-2x)^2 = 3240x^2\)- For \(k=3\): \( \binom{6}{3} (3)^3 (-2x)^3 = -4320x^3\).
04

Combine the Terms

Combine the terms calculated:\[(3-2x)^6 = 729 - 1458x + 3240x^2 - 4320x^3 + \ldots\] The terms beyond \(x^3\) are not needed for this problem.

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Key Concepts

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

Binomial Coefficients
In mathematics, binomial coefficients are a fundamental concept used to determine the number of ways to choose a subset of items from a larger set. These coefficients are denoted as \( \binom{n}{k} \), which represents the number of ways to choose \( k \) elements from a set of \( n \) elements without regard to the order of selection. This is also known as "n choose k."
  • The formula for calculating a binomial coefficient is \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \( n! \) (n factorial) is the product of all positive integers up to \( n \).
  • A practical example would be \( \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15 \).
Binomial coefficients are essential in the expansion of binomials using the binomial theorem. Each coefficient in the expanded form corresponds to a term with the respective power of the variables.
Polynomial Expansion
Polynomial expansion, particularly with the help of the binomial theorem, makes it straightforward to raise a binomial expression to a power accurately and efficiently. This process transforms expressions like \((3-2x)^6\) into a sum of terms of descending powers.
  • Each term in the expansion is computed by multiplying a binomial coefficient with powers of the original terms \(a\) and \(b\), as shown in \( \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \).
  • In this example, \(a\) is 3 and \(b\) is \(-2x\), meaning each term results from pushing \( (3)^{6-k} \) and \((-2x)^k \) through the binomial coefficient \( \binom{6}{k} \).
This method allows the determination of specific terms, such as up to \(x^3\), without expanding the entire binomial power.
Mathematical Sequences
Mathematical sequences are ordered lists of numbers following a specific pattern or rule. In the context of polynomial expansion processes, these sequences can help organize computations systematically.
  • When expanding using the binomial theorem, the individual terms follow a sequence determined by the binomial coefficients. For \((3-2x)^6\), this sequence of coefficients is 1, 6, 15, 20, 15, 6, 1.
  • The numbers correspond to the successive powers of 3 and \(-2x\), forming terms like \(729\), \(-1458x\), \(3240x^2\), and so on, maintaining an ordered sequence in the adjusted polynomial expression.
Recognizing these sequences assists in identifying patterns that help simplify complex calculations and provides a deeper understanding of the polynomial's structure.

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Most popular questions from this chapter

The sum to 20 terms of an arithmetic series is identical to the sum to 22 terms. If the common difference is \(-2\), find the first term.

(a) State the power series expansion for \(\mathrm{e}^{-x}\). (b) By using your solution to (a) and the expansion for \(e^{x}\), deduce the power series expansions of \(\cosh x\) and \(\sinh x\).

A geometric progression has first term \(a=1\). The ninth term exceeds the fifth tem by 240 . Find possible values for the eighth term.

If the sum to 10 terms of an arithmetic series is 100 and its common difference, \(d\), is \(-3\), find its first term.

Obtain the first five terms in the expansion of \((1+2 x)^{1 / 2}\). State the range of values of \(x\) for which the expansion is valid. Choose a value of \(x\) within the range of validity and compute values of your expansion for comparison with the true function values.
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