Can you prove the summation 13+23+33+...+n3=(1+2+...n)2?

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Discussion Overview

The discussion centers around proving the summation formula for the cubes of the first n natural numbers, specifically that \( 1^3 + 2^3 + 3^3 + \ldots + n^3 = (1 + 2 + \ldots + n)^2 \). Participants explore various approaches and references related to this mathematical identity, including numerical observations and references to established formulas.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses a numerical understanding of the identity but seeks an analytical proof, suggesting induction as a potential method.
  • Another participant references Faulhaber's formula, indicating that it may relate to the problem, particularly for the case where \( p=3 \).
  • A third participant points to the concept of squared triangular numbers as potentially relevant to the discussion.
  • One participant notes a lack of derivation for the identity and contrasts it with Faulhaber's formula, questioning the application of \( p=3 \) on both sides of the equation.
  • Another participant mentions a proof related to the sum of consecutive odd numbers, suggesting a connection to the identity being discussed.
  • A more technical approach is proposed involving the derivation of closed forms for sums of powers, suggesting a method to isolate the case for \( p=n \) using known closed forms for lower powers.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the proof or derivation of the identity. Multiple approaches and references are presented, indicating a variety of perspectives and unresolved questions regarding the proof.

Contextual Notes

Some participants reference established mathematical concepts and theorems, but there is uncertainty regarding the applicability of these references to the specific identity in question. The discussion includes assumptions about the validity of various mathematical approaches without full resolution.

Loren Booda
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Given nonzero whole numbers n, prove

13+23+33+...+n3=(1+2+...n)2

I figured this out numerically, but lack the skills to solve it analytically (no doubt by induction) and could not find it in my table of summations. I'm too old for this to be homework.
 
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Loren Booda said:
Given nonzero whole numbers n, prove

13+23+33+...+n3=(1+2+...n)2

I figured this out numerically, but lack the skills to solve it analytically (no doubt by induction) and could not find it in my table of summations. I'm too old for this to be homework.

http://en.wikipedia.org/wiki/Faulhaber's_formula

look at the case p=3
 
Thanks, ice109. I'm surprised that I've never heard of Faulhaber before. I guess Bernoulli got all the acclaim.

I still don't see a derivation of my finding, however. I thought in the case I presented that p=3 on one side of the equation, and p=2 on the other, as opposed to p=3 for both sides of Faulhaber's formula.

K.J.Healey seems to have what I seek, including a proof of Nicomachus's theorem.
 
i like that one better

what the hell? every n^3 is the sum of n consecutive odd numbers? wheatstone's proof seems to imply that
 
Last edited:
If one knows the closed forms for all cases p< n, then the closed form for p=n can be derived as such:

Set up a table into two columns, LHS and RHS of the following equation;

[tex](x+1)^{n+1} - x^{n+1} = (^{n+1}C_1)x^n + (^{n+1}C_2)x^{n-1} ...+1[/tex].

Sum this expression for k=1, 2,3,4...m. The LHS is a telescoping series. The RHS is the sum of cases p=0, 1, 2, 3... n. Replace every series with its known closed form. Then isolate the p=n case onto one side of the equation, and simplify.

Since we only need this for up to p=3, it shouldn't be very hard.
 

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