How do we find the term #n^(th)# and sum to #n^(th)# of the following series?
I won't derive these common sums:
The sum of the squares is
The sum of cubes is remarkably the square of the triangular numbers, the sums of the natural numbers.
Is that a second sum? I'll leave that to others.
Let's do the second sum.
The inner sum is known:
The sum of cubes is the square of the triangular number,
The triangular numbers are the sums of the naturals,