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

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In summary, the closed form for the summation of consecutive cubes starting from 1 up to n can be derived using the closed forms for cases p=0, 1, and 2 and the telescoping series method.
  • #1
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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  • #2
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
 
  • #4
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.
 
  • #5
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
 
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  • #6
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.
 

1. What is the formula for the summation 13+23+33+...+n3?

The formula for this summation is n(n+1)^2/4.

2. Can you explain how this formula was derived?

This formula can be derived using mathematical induction and the fact that the sum of the first n cubes is equal to (n(n+1)/2)^2.

3. How can you prove that n(n+1)^2/4 = (1+2+...n)2?

The proof involves using the formula for the sum of the first n cubes and simplifying the expression using algebraic manipulations.

4. Is this formula applicable for all values of n?

Yes, this formula is applicable for all positive integer values of n.

5. Are there any real-life applications of this summation formula?

Yes, this summation formula is commonly used in the field of calculus to solve various problems related to sequences and series.

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