Natural Numbers and Induction (Analysis with and Introduction to ProofC)

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Homework Help Overview

The problem involves proving the equality of the sum of cubes of natural numbers and the square of the sum of the first n natural numbers, specifically stated as 1^3 + 2^3 + ... + n^3 = (1 + 2 + ... + n)^2 for all natural numbers n. The context is centered around mathematical induction as a proof technique.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to use mathematical induction, starting with the base case and assuming the statement holds for k, then considering k+1. Some participants suggest simplifying the sum of the first k natural numbers and provide formulas related to the sums involved.

Discussion Status

The discussion is ongoing, with the original poster expressing uncertainty about the next steps in their proof. Some participants have provided insights and formulas that may assist in the proof, but there is no explicit consensus on the approach or resolution yet.

Contextual Notes

Participants are discussing the proof within the constraints of natural numbers and the rules of mathematical induction. There is an acknowledgment of the formulas for the sums involved, but the original poster has not yet fully resolved their proof attempt.

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Homework Statement



Prove that: 1^3 + 2^3 + ... + n^3 = (1 + 2 + ... + n)^2 for all n where n is a natural number

Homework Equations



Proof by induction:
a) p(1) is true
b) assume p(k) is true then prove p(k+1) for it

The Attempt at a Solution



I gave this a try for a while but I'm quite stuck.

So far using proof by induction:
p(1) = 1^3 = 1^2 is true

assume that p(k) = 1^3 + 2^3 + ... + k^3 = (1 + 2 + ... + k)^2 is true
then:
p(k+1) = 1^3 + 2^3 + ... + k^3 + (k + 1)^3
= (1 + 2 + ... + k)^2 + (k + 1)^3

Now I'm really not sure where to go or if I'm going the right way. Any sort of hint would be greatly appreciated. Thanks.

-Kentt
 
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Simplify 1+2+3+...k.
 
[tex]\sum_{n=1}^k n= \frac{(k+1)(k)}{2}[/tex] btw.
You may also like to know that [tex]\sum_{n=1}^k n^3 = \frac{(k+1)^2k^2}{4}=(\frac{(k+1)k}{2})^2[/tex]...o nvm Thats what your trying to prove...
 
Thanks IMDerek. Got it done.
 

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