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Sum of Cubes (Posted on 2004-05-25) Difficulty: 3 of 5
Prove that the sum of consecutive perfect cubes (starting with 1) is always a perfect square.

For example:
1=1
1+8=9
1+8+27=36

See The Solution Submitted by Gamer    
Rating: 3.4000 (5 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution algebraic solution | Comment 3 of 11 |
1=1                                           1
1+8=9                                       3
1+8+27=36                                6
1+8+27+64=100                       10
1+8+27+64+125=225               15
1+8+27+64+125+216=441        21

for ex. 6 + (tr(5)) = (tr(6))

given :tr(x) = (x+x)/2
           tr(x-1) = (x-x)/2

prove: n+(tr(n-1)) = (tr(n))

n + ((n-n)/2) = (n+n)/2)
n + (.5n - .5n) = (.5n + .5n)
n + .25n^4 - .5n + .25n = .25n^4 + .5n + .25n
n - .5n + .25n = .5n + .25n
n - .5n = .5n
n = n

Edited on May 25, 2004, 4:48 pm
  Posted by Danny on 2004-05-25 16:45:17

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