n e induction on 17 to prove the Binomial Theorem (a + b)" = Σ C(n, k) an-kbk for all positive integers n ≥ 1.
n e induction on 17 to prove the Binomial Theorem (a + b)" = Σ C(n, k) an-kbk for all positive integers n ≥ 1.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Title: Inductive Proof of the Binomial Theorem**
**Objective:**
To prove the Binomial Theorem using mathematical induction for all positive integers \( n \geq 1 \).
**Statement of the Binomial Theorem:**
The theorem states that for any positive integer \( n \), the expansion of \( (a+b)^n \) can be expressed as:
\[
(a+b)^n = \sum_{k=0}^{n} C(n, k) \cdot a^{n-k} \cdot b^k
\]
where \( C(n, k) \) is the binomial coefficient, calculated as:
\[
C(n, k) = \frac{n!}{k!(n-k)!}
\]
**Explanation:**
The formula gives the expansion of the binomial expression \( (a+b)^n \) into a sum involving terms of the form \( C(n, k) \cdot a^{n-k} \cdot b^k \). Each term in this expansion consists of a coefficient \( C(n, k) \), which represents the number of ways to choose \( k \) elements from a set of \( n \), multiplied by powers of \( a \) and \( b \).
**Method of Induction:**
To prove this theorem by induction, follow these steps:
1. **Base Case:**
Verify the theorem for the initial value of \( n \) (often \( n = 1 \)).
2. **Inductive Step:**
Assume the theorem holds for \( n = m \), i.e.,
\[
(a+b)^m = \sum_{k=0}^{m} C(m, k) \cdot a^{m-k} \cdot b^k
\]
Show that if the theorem holds for \( n = m \), then it must also hold for \( n = m+1 \):
\[
(a+b)^{m+1} = \sum_{k=0}^{m+1} C(m+1, k) \cdot a^{m+1-k} \cdot b^k
\]
By successfully completing these steps, the theorem is proven for all positive integers \( n \geq 1 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F11b3ac0d-2603-4899-810b-1b144399767c%2F969e7ce9-9006-4dad-a599-4ebe3968e479%2Fntzjgtt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Inductive Proof of the Binomial Theorem**
**Objective:**
To prove the Binomial Theorem using mathematical induction for all positive integers \( n \geq 1 \).
**Statement of the Binomial Theorem:**
The theorem states that for any positive integer \( n \), the expansion of \( (a+b)^n \) can be expressed as:
\[
(a+b)^n = \sum_{k=0}^{n} C(n, k) \cdot a^{n-k} \cdot b^k
\]
where \( C(n, k) \) is the binomial coefficient, calculated as:
\[
C(n, k) = \frac{n!}{k!(n-k)!}
\]
**Explanation:**
The formula gives the expansion of the binomial expression \( (a+b)^n \) into a sum involving terms of the form \( C(n, k) \cdot a^{n-k} \cdot b^k \). Each term in this expansion consists of a coefficient \( C(n, k) \), which represents the number of ways to choose \( k \) elements from a set of \( n \), multiplied by powers of \( a \) and \( b \).
**Method of Induction:**
To prove this theorem by induction, follow these steps:
1. **Base Case:**
Verify the theorem for the initial value of \( n \) (often \( n = 1 \)).
2. **Inductive Step:**
Assume the theorem holds for \( n = m \), i.e.,
\[
(a+b)^m = \sum_{k=0}^{m} C(m, k) \cdot a^{m-k} \cdot b^k
\]
Show that if the theorem holds for \( n = m \), then it must also hold for \( n = m+1 \):
\[
(a+b)^{m+1} = \sum_{k=0}^{m+1} C(m+1, k) \cdot a^{m+1-k} \cdot b^k
\]
By successfully completing these steps, the theorem is proven for all positive integers \( n \geq 1 \).
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