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
We observe that:
Sum( i^4-(i-1)^4 ) = 1^4-0^4+2^4-1^4+....+(n-1)^4-(n-2)^4+n^4-(n-1)^4
i= 1 to n
= n^4
Also,:
Sum (i^4-(i-1)^4)
i=1 to n
=4*S(3)-6*S(2)+4*S(1)-n, where:
Sum (i^n) = S (n)
i=1 to n
Therefore, we must have:
n^4=4*S(3)-6*S(2)+4*S(1)-n
=> 4S(3)
= n^4+n(n+1)(2n+1)-2n(n+1)+n
= n^4+n+n(n+1)(2n+1)-2n(n+1)
=n(n+1)(n^2-n+1+2n+1-2)
=n(n+1)(n^2+n)
= n^2*(n+1)^2
Accordingly,
S(3) = (n(n+1)/2)^2
Therefore,
Sum (i^3) ={n(n+1)/2}^2
i=1 to n
Consequently, the sum of consecutive perfect cubes is a perfect square.
Edited on January 6, 2022, 1:54 am
Puzzle Solution: Formal Method
