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

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SUMMARY

The forum discussion centers on proving the equation 1^3 + 2^3 + ... + n^3 = (1 + 2 + ... + n)^2 for all natural numbers n using mathematical induction. The proof involves verifying the base case p(1) and assuming the statement holds for p(k) before proving it for p(k+1). The user Kentt initially struggles with the proof but receives guidance from IMDerek, who simplifies the summation of natural numbers and confirms the correctness of the approach.

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  • Understanding of mathematical induction
  • Familiarity with summation formulas, specifically for natural numbers
  • Knowledge of cubic functions and their properties
  • Basic algebraic manipulation skills
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  • Learn about summation formulas for cubes and their derivations
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Students of mathematics, educators teaching proof techniques, and anyone interested in the fundamentals of mathematical induction and summation formulas.

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

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