sigma(n)^2 == sigma(n^3)

Let's create a list of the first n positive integers:

Now let's add them up and square the sum.

In other words, we are finding (1 + 2 + 3 + ... + n)^2.

Expressed in sigma notation, we have just found (\sum_{x = 1}^{n}x)^2.

Now let's find the cubes of the original n terms.

And now let's add up the cubes.  In other words, we are finding 1^3 + 2^3 + 3^3 + ... + n^3.

Expressed in sigma notation, we have found \sum_{x = 1}^{n}x^3

Experiment with different values of n and verify whether or not

(1 + 2 + 3 + ... + n)^2 = 1^3 + 2^3 + 3^3 + ... + n^3,

or (\sum_{x = 1}^{n}x)^2 = \sum_{x = 1}^{n}x^3.