Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Question:**
When multiplied out, what are the first 4 terms of \((2x + y)^{10}\)?
Show your work for full credit.
---
**Explanation:**
To solve this problem, we will use 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 given by \(\frac{n!}{k!(n-k)!}\).
In this specific case, \(a = 2x\), \(b = y\), and \(n = 10\).
The first 4 terms of \((2x + y)^{10}\) are obtained by setting \(k = 0\), \(k = 1\), \(k = 2\), and \(k = 3\).
1. For \(k = 0\):
\[
\binom{10}{0} (2x)^{10} y^0 = 1 \cdot (2x)^{10} \cdot 1 = 1024x^{10}
\]
2. For \(k = 1\):
\[
\binom{10}{1} (2x)^9 y^1 = 10 \cdot (2x)^9 \cdot y = 10 \cdot 512x^9 \cdot y = 5120x^9y
\]
3. For \(k = 2\):
\[
\binom{10}{2} (2x)^8 y^2 = 45 \cdot (2x)^8 \cdot y^2 = 45 \cdot 256x^8 \cdot y^2 = 11520x^8y^2
\]
4. For \(k = 3\):
\[
\binom{10}{3} (2x)^7 y^3 = 120 \cdot (2x)^7 \cdot y^3 = 120 \cdot 128x^7 \cdot y^3 = 15360x^7y^3
\]
Thus, the first 4 terms of \((2x + y)^{10](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8cf7e672-b05a-4acb-ad51-f9b37172b8f4%2F0fec02e9-db5a-4394-af34-1c0840fa67de%2Ft4jv5i_processed.png&w=3840&q=75)
Transcribed Image Text:**Question:**
When multiplied out, what are the first 4 terms of \((2x + y)^{10}\)?
Show your work for full credit.
---
**Explanation:**
To solve this problem, we will use 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 given by \(\frac{n!}{k!(n-k)!}\).
In this specific case, \(a = 2x\), \(b = y\), and \(n = 10\).
The first 4 terms of \((2x + y)^{10}\) are obtained by setting \(k = 0\), \(k = 1\), \(k = 2\), and \(k = 3\).
1. For \(k = 0\):
\[
\binom{10}{0} (2x)^{10} y^0 = 1 \cdot (2x)^{10} \cdot 1 = 1024x^{10}
\]
2. For \(k = 1\):
\[
\binom{10}{1} (2x)^9 y^1 = 10 \cdot (2x)^9 \cdot y = 10 \cdot 512x^9 \cdot y = 5120x^9y
\]
3. For \(k = 2\):
\[
\binom{10}{2} (2x)^8 y^2 = 45 \cdot (2x)^8 \cdot y^2 = 45 \cdot 256x^8 \cdot y^2 = 11520x^8y^2
\]
4. For \(k = 3\):
\[
\binom{10}{3} (2x)^7 y^3 = 120 \cdot (2x)^7 \cdot y^3 = 120 \cdot 128x^7 \cdot y^3 = 15360x^7y^3
\]
Thus, the first 4 terms of \((2x + y)^{10
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