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Sum of First n Cubes, First n Squares

Date: 11/18/2002 at 20:48:50
From: Sam
Subject: Math Team Test Question

I have a question for my math team at school.  Is there a shortcut to 
find (1^3-1^2)+(2^3-2^2)+(3^3-3^2)... (15^3-15^2) ? 

Thanks!


Date: 11/18/2002 at 21:20:48
From: Doctor Paul
Subject: Re: Math Team Test Question

Rearrange the parentheses to obtain:

   1^3 + 2^3 + ... + 15^3 - (1^2 + 2^2 + ... + 15^2)

There are formulas for the sum of the first n cubes and the sum of 
the first n squares. In fact, there is a formula for the sum of the 
first n kth powers for any positive integer k.

The sum of the first n cubes is given by:

   1/4*n^2*(n+1)^2

The sum of the first n squares is given by:

   1/6*n*(n+1)*(2*n+1)

The difference of these two expressions is given by:

   1/4*n^4 + 1/6*n^3 - 1/4*n^2 - 1/6*n

we want to evaluate the above expression at n = 15.

I get 13160.

If it seems as if I just pulled the above formulas out of thin air, 
I'd encourage you to read some of the following from our archives:

   Finding sum formula using sequences of differences
   http://mathforum.org/library/drmath/view/56383.html 

   Formula for Sum of First N squares
   http://mathforum.org/library/drmath/view/56988.html 

   Formula For the Sum Of the First N Squares
   http://mathforum.org/library/drmath/view/56920.html 

I hope this helps.  Please write back if you'd like to talk about 
this some more.

- Doctor Paul, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Exponents
High School Number Theory

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