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Chapter 14: Problem 34

Use the binomial theorem to expand each binomial. $$ \left(y^{3}+2\right)^{4} $$

Short Answer

Expert verified
\( y^{12} + 8y^9 + 24y^6 + 32y^3 + 16 \)

Step by step solution

01

- Write down the Binomial Theorem

The Binomial Theorem formula is given by \( (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k} \). This will be used to expand the given binomial.
02

- Identify the values of a, b, and n

In this exercise, \(a = y^3\), \(b = 2\), and \(n = 4\).
03

- Set up the expansion using the Binomial Theorem

Substitute the identified values into the Binomial Theorem formula: \( (y^3 + 2)^4 = \sum_{k=0}^{4} \binom{4}{k} (y^3)^{4-k} (2)^k \).
04

- Calculate each term

Compute each term in the expansion: \( \binom{4}{0}(y^3)^{4-0}(2)^0 + \binom{4}{1}(y^3)^{4-1}(2)^1 + \binom{4}{2}(y^3)^{4-2}(2)^2 + \binom{4}{3}(y^3)^{4-3}(2)^3 + \binom{4}{4}(y^3)^{4-4}(2)^4 \).
05

- Simplify each term

Simplify each of the binomial coefficients and powers: \( \binom{4}{0}(y^3)^4(2)^0 = 1 \cdot y^{12} \cdot 1 = y^{12} \), \( \binom{4}{1}(y^3)^3(2)^1 = 4 \cdot y^9 \cdot 2 = 8y^9 \), \( \binom{4}{2}(y^3)^2(2)^2 = 6 \cdot y^6 \cdot 4 = 24y^6 \), \( \binom{4}{3}(y^3)^1(2)^3 = 4 \cdot y^3 \cdot 8 = 32y^3 \), \( \binom{4}{4}(y^3)^0(2)^4 = 1 \cdot 1 \cdot 16 = 16 \).
06

- Summarize the expansion

Combine all simplified terms to write the complete expansion: \( y^{12} + 8y^9 + 24y^6 + 32y^3 + 16 \).

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Key Concepts

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

Binomial Expansion
Binomial expansion allows us to express a binomial raised to a power as a sum of terms. This is incredibly useful for simplifying expressions or solving equations. The general form relies on the Binomial Theorem: \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}\). Here, each term involves a combination of the components \(a\) and \(b\) raised to specific powers and multiplied by a binomial coefficient.
Binomial Coefficients
Binomial coefficients are the numbers that appear as multipliers in the terms of a binomial expansion. They can be calculated using the combination formula: \( \binom{n}{k} = \frac{n!}{k! (n-k)!} \). They represent the number of ways to choose \(k\) elements from a set of \(n\) elements. For example, in the expansion of \((y^3 + 2)^4\), the binomial coefficients are \(\binom{4}{0}, \binom{4}{1}, \binom{4}{2}, \binom{4}{3}, \binom{4}{4}\), which simplify to 1, 4, 6, 4, and 1 respectively.
Powers of Binomials
When dealing with binomials raised to a power, the goal is to expand the expression from a condensed form. For instance, in \((y^3 + 2)^4\), you will determine each term's power using the formula \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}\). By breaking this down:
  • The first term \(a\) starts with the highest power \(n\) and decreases.
  • The second term \(b\) starts with zero and increases.
For the term involving \(y^3\) and 2, each power adjustment must be performed until all terms are calculated. Combining these, we get the expansion \( y^{12} + 8y^9 + 24y^6 + 32y^3 + 16 \). This detailed breakdown helps in understanding how binomial expansions work.

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Most popular questions from this chapter

A student incorrectly claimed that the common difference for the arithmetic sequence $$ -15,-10,-5,0,5, \ldots $$ is - 5. WHAT WENT WRONG? Find the correct common difference.

Determine an expression for the general term of each arithmetic sequence. Then find \(a_{25}\). \(-10,-5,0, \ldots\)

Find the sum, if it exists, of the terms of each infinite geometric sequence. See Examples 8 and 9. $$ \sum_{i=1}^{\infty} \frac{3}{5}\left(\frac{5}{6}\right)^{i} $$

Determine an expression for the general term of each arithmetic sequence. Then find \(a_{25}\). \(1, \frac{5}{3}, \frac{7}{3}, 3, \ldots\)

If the given sequence is geometric, find the common ratio \(r .\) If the sequence is not geometric, say so. See Example 1. $$ \frac{2}{3},-\frac{2}{15}, \frac{2}{75},-\frac{2}{375}, \ldots $$
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