Sum of Cubed Integers
This will simply show how to sum all cubed integers, 1 + 8 + 27 ... + n3, let this be S3n. Let the sum of integers be Sn and the sum of squared integers be S2n
Take the expression
(x - 1)4 = x4 - 4x3 + 6x2 - 4x + 1
This can be re-arranged to
x4 - (x - 1)4 = 4x3 - 6x2 + 4x - 1
Now if we sum each side from 1 to n, all the terms on the left would cancel leaving just n4
n4 - (n - 1)4 + (n - 1)4 - (n - 2)4 + (n - 2)4 - (n - 3)4 + .......... + 1 - 0
Hencen4 = 4S3n - 6S2n + 4Sn - n
We already know that Sn = n/2 (n + 1) and S2n = 1/6n(n + 1)(2n + 1), therefore
4S3n = n4 + n(n + 1)(2n + 1) - 2n(n + 1) + n
Hence4S3n = n(n3 + (n + 1)(2n + 1) - 2(n + 1) + 1) = n2(n + 1)2
ThereforeS3n = [n/2(n + 1)]2
Resulting in the expression
S3n = Sn2
This is very interesting as it proves that this sum will always be a perfect square.
Updated: 10/12/2023