Proving Induction and Divisibility - Two Simple Homework Problems

  • Thread starter Thread starter steelphantom
  • Start date Start date
  • Tags Tags
    Induction Proofs
Click For Summary

Homework Help Overview

The discussion revolves around two mathematical proofs involving induction and divisibility. The first problem asks participants to prove that the sum of cubes of the first n natural numbers equals the square of the sum of the first n natural numbers. The second problem involves proving that a specific expression involving powers of 7 and linear terms is divisible by 36 for all positive integers n.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the typical structure of induction proofs, with attempts to establish a base case and an inductive step for both problems. Questions arise about how to manipulate the expressions to reach the desired forms, particularly regarding the sum of integers and its relationship to the sum of cubes.

Discussion Status

Some participants have made progress on the second problem, indicating they have solved it, while the first problem remains a point of confusion for others. Guidance has been offered regarding the use of closed forms for sums, and there is ongoing exploration of how to combine terms in the first problem.

Contextual Notes

Participants are encouraged to derive formulas and explore algebraic manipulations, with some expressing uncertainty about specific steps in their reasoning. There is a focus on understanding the implications of the assumptions made in the proofs.

steelphantom
Messages
158
Reaction score
0

Homework Statement


Prove 1^3 + 2^3 + ... + n^3 = (1 + 2 + ... + n)^2 for all natural numbers n.


Homework Equations





The Attempt at a Solution



Well, this seems like the typical induction proof, so I start by testing the hypothesis at 1: 1^3 = 1^2 = 1. Then I assume that the equation is true for n, and try to prove it true for n + 1.

1 ^3 + 2^3 + ... + n^3 + (n + 1)^3 = (1 + 2 + ... + n)^2 + (n + 1)^3

I have this, but how do I get it to be (1 + 2 + ... + n + n + 1)^2? Is it some crazy factorization that I'm overlooking? It seems pretty easy.


Homework Statement



Prove that 7^n - 6n - 1 is divisible by 36 for all positive integers n.

Homework Equations





The Attempt at a Solution



Well, the statement seems to be saying the following: 7^n - 6n - 1 = 36k, where k is a positive integer. As with the first problem, testing it with 1 works. Assuming it is true for n + 1 would mean the following, if I'm correct:

7^(n + 1) - 6(n + 1) - 1 = 36j, where j is a positive integer. At this point, I'm not sure how to proceed.


I know I'm probably overlooking some obvious stuff here, so a little hint is probably all I'll need to finish these problems off. Thanks! :)
 
Physics news on Phys.org
steelphantom said:

Homework Statement


Prove 1^3 + 2^3 + ... + n^3 = (1 + 2 + ... + n)^2 for all natural numbers n.


Homework Equations





The Attempt at a Solution



Well, this seems like the typical induction proof, so I start by testing the hypothesis at 1: 1^3 = 1^2 = 1. Then I assume that the equation is true for n, and try to prove it true for n + 1.

1 ^3 + 2^3 + ... + n^3 + (n + 1)^3 = (1 + 2 + ... + n)^2 + (n + 1)^3

I have this, but how do I get it to be (1 + 2 + ... + n + n + 1)^2? Is it some crazy factorization that I'm overlooking? It seems pretty easy.
Try to substitute something else for (1+2+...+n). The formula for it is not that hard to derive.
steelphantom said:

Homework Statement



Prove that 7^n - 6n - 1 is divisible by 36 for all positive integers n.

Well, the statement seems to be saying the following: 7^n - 6n - 1 = 36k, where k is a positive integer. As with the first problem, testing it with 1 works. Assuming it is true for n + 1 would mean the following, if I'm correct:

7^(n + 1) - 6(n + 1) - 1 = 36j, where j is a positive integer. At this point, I'm not sure how to proceed.
7^(n+1) = 7^n*7. You are assuming that 7^n = 36k + 6n + 1. See what happens.
 
steelphantom said:

Homework Statement


Prove 1^3 + 2^3 + ... + n^3 = (1 + 2 + ... + n)^2 for all natural numbers n.

1^3 + 2^3 + ... + n^3=\sum_{n=0} ^{N} n^3

(1 + 2 + ... + n)^2 =(\sum_{n=0} ^{N} n)^2


Do you know what \sum_{n=0} ^{N} n represents?
 
Thanks guys for your help! I solved the second problem. The first problem is still puzzling me a bit. Here's what I have:

1^3 + 2^3 + ... + n^3 = \sum_{n=0} ^{N} n)^2 + (n + 1)^3

Still not sure how to add those two terms together. I'm sure it's very easy, but I'm just not seeing it.
 
The same question, again, can you write
\sum_{n=0}^N n
in a "closed" form and then square?
 
HallsofIvy said:
The same question, again, can you write
\sum_{n=0}^N n
in a "closed" form and then square?

Got it! Thank you. :)
 

Similar threads

Replies
6
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K