Follow-Ups to Answer from Expert Paul Klarreich

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William wrote at 2007-06-14 23:11:54
Use (1+n)^3 = 1 +  3n    +  3n^2    +  n^3  

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

         .

      2^3  = 1 + 3      + 3        + 1



Sum up =>



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

         +(1+...+n^3)



cancel some terms out =>

-1+(1+n)^3=n+3*n(n+1)/2+3(1+...+n^2)

then the rest is easy work out (1+...+n^2)



Same way using (1+n)^4 can work out

    (1+...+n^3)  

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Paul Klarreich wrote at 2007-07-07 03:38:16
Indeed, this is a very nice approach.  I can't remember ever having seen it before.  It works out as shown below.  I felt it necessary to work out all the algebra just to see if I can still do it, and having done it, I might as well send it along.



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

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

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

...

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

 1^3      = 1 + 3(0)   + 3(0)     + 0^3

-------------------------------------------------

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



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



Now the sum of cubes will cancel:



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

And this is what we seek:          ***************



The sum of 1..n can be rewritten:



(n+1)^3 = n+1 + 3(n(n + 1)/2) + 3(SUM OF SQUARES)



Now some algebra:



n^3 + 3n^2 + 3n + 1 - n - 1 - 3(n(n + 1)/2) = 3(SUM OF SQUARES)



n^3 + 3n^2 + 2n  - 3(n^2 + n)/2) = 3(SUM OF SQUARES)



[2n^3 + 6n^2 + 4n  - 3n^2 - 3n)]/2 = 3(SUM OF SQUARES)



[2n^3 + 3n^2 + n ]/2 = 3(SUM OF SQUARES)



[2n^3 + 3n^2 + n ]/6 = (SUM OF SQUARES)



[n(2n^2 + 3n + 1]/6 = (SUM OF SQUARES)



[n(2n + 1)(n + 1)]/6 = (SUM OF SQUARES)



and that is the formula.

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Karoye wrote at 2010-08-25 18:32:06
Esta fórmula es el resultado del método de las diferencias finitas, el cual es a la vez la comprobación de la misma.

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