7. Prove the following using induction. n Σk³ = 1³ + 2³ +3³ + ... + n³ = (1 + 2 + 3 + ... + n)² 13 23 k=1 .. [Hint: Recall that 1 + 2 + 3 + ··· + n = n(n+1) 2 ]
7. Prove the following using induction. n Σk³ = 1³ + 2³ +3³ + ... + n³ = (1 + 2 + 3 + ... + n)² 13 23 k=1 .. [Hint: Recall that 1 + 2 + 3 + ··· + n = n(n+1) 2 ]
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Discrete Mathematics
![### Inductive Proof of Sum of Cubes Formula
**Problem:**
Prove the following using induction.
\[
\sum_{k=1}^{n} k^3 = 1^3 + 2^3 + 3^3 + \cdots + n^3 = \left(1 + 2 + 3 + \cdots + n\right)^2
\]
**Hint:**
Recall that \(1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}\).
### Explanation of the Formula and Steps for Inductive Proof:
1. **Base Case:**
For \(n = 1\):
\[
\sum_{k=1}^{1} k^3 = 1^3 = 1
\]
\[
\left(1 + 2 + 3 + \cdots + 1\right)^2 = 1^2 = 1
\]
The base case holds true since both sides are equal to 1.
2. **Inductive Step:**
Assume the formula holds for some integer \(n = k\), i.e.,
\[
\sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2
\]
Now show that the formula holds for \(n = k + 1\):
Consider the sum \(\sum_{k=1}^{n+1} k^3\):
\[
\sum_{k=1}^{n+1} k^3 = \sum_{k=1}^{n} k^3 + (n+1)^3
\]
Using the inductive hypothesis:
\[
\sum_{k=1}^{n+1} k^3 = \left( \frac{n(n+1)}{2} \right)^2 + (n+1)^3
\]
Simplify and show that this equals \(\left( \frac{(n+1)(n+2)}{2} \right)^2\).
This proof technique ensures that the stated formula \(\sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2\) is](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F11ae144f-50cb-4873-b91f-81da3ae321d6%2F7dbaac9a-6bd0-4627-844c-f4be21a2cf94%2Fdtcv2d_processed.png&w=3840&q=75)
Transcribed Image Text:### Inductive Proof of Sum of Cubes Formula
**Problem:**
Prove the following using induction.
\[
\sum_{k=1}^{n} k^3 = 1^3 + 2^3 + 3^3 + \cdots + n^3 = \left(1 + 2 + 3 + \cdots + n\right)^2
\]
**Hint:**
Recall that \(1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}\).
### Explanation of the Formula and Steps for Inductive Proof:
1. **Base Case:**
For \(n = 1\):
\[
\sum_{k=1}^{1} k^3 = 1^3 = 1
\]
\[
\left(1 + 2 + 3 + \cdots + 1\right)^2 = 1^2 = 1
\]
The base case holds true since both sides are equal to 1.
2. **Inductive Step:**
Assume the formula holds for some integer \(n = k\), i.e.,
\[
\sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2
\]
Now show that the formula holds for \(n = k + 1\):
Consider the sum \(\sum_{k=1}^{n+1} k^3\):
\[
\sum_{k=1}^{n+1} k^3 = \sum_{k=1}^{n} k^3 + (n+1)^3
\]
Using the inductive hypothesis:
\[
\sum_{k=1}^{n+1} k^3 = \left( \frac{n(n+1)}{2} \right)^2 + (n+1)^3
\]
Simplify and show that this equals \(\left( \frac{(n+1)(n+2)}{2} \right)^2\).
This proof technique ensures that the stated formula \(\sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2\) is
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps with 2 images
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
