#### Sum of Cubed Integers

This will simply show how to sum all cubed integers, 1 + 8 + 27 ... + n

^{3}, let this be S_{3n}. Let the sum of integers be S_{n}and the sum of squared integers be S_{2n}Take the expression(x - 1)

This can be re-arranged to
^{4}= x^{4}- 4x^{3}+ 6x^{2}- 4x + 1x

Now if we sum each side from 1 to n, all the terms on the left would cancel leaving just n^{4}- (x - 1)^{4}= 4x^{3}- 6x^{2}+ 4x - 1^{4}n

Hence^{4}- (n - 1)^{4}+ (n - 1)^{4}- (n - 2)^{4}+ (n - 2)^{4}- (n - 3)^{4}+ .......... + 1 - 0n

We already know that S^{4}= 4S_{3n}- 6S_{2n}+ 4S_{n}- n_{n}= n/2 (n + 1) and S_{2n}= 1/6n(n + 1)(2n + 1), therefore4S

Hence_{3n}= n^{4}+ n(n + 1)(2n + 1) - 2n(n + 1) + n4S

Therefore_{3n}= n(n^{3}+ (n + 1)(2n + 1) - 2(n + 1) + 1) = n^{2}(n + 1)^{2}S

Resulting in the expression
_{3n}= [n/2(n + 1)]^{2}S

This is very interesting as it proves that this sum will always be a perfect square.
_{3n}= S_{n}^{2}Updated: 10/12/2023