Sum of First n Cubes, First n SquaresDate: 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/ |
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