To prove this equation, we will utilize mathematical induction.

First, let's prove that the equation holds for n = 1.

When n = 1:

1(3*1+1) = 1(1+1)^2
4 = 4

Now, let's assume that the equation holds for some positive integer k, where k > 1. Therefore, the sum up to k terms can be expressed as:

1(31+1) + 2(32+1) + 3(3*3+1) + ... + k(3k+1) = k(k+1)^2

Next, let's prove that the equation holds for k+1.

Adding the (k+1)th term to both sides of the equation gives:

1(31+1) + 2(32+1) + 3(3*3+1) + ... + k(3k+1) + (k+1)(3(k+1)+1) = (k+1)((k+1)+1)^2

Now, simplify the left side:

k(k+1)^2 + (k+1)(3k+4) = (k+1)(k+2)^2
k(k+1)^2 + 3k^2 + 4k + 3k + 4 = (k+1)(k+2)^2
k(k+1)^2 + 3k^2 + 7k + 4 = (k+1)(k^2 + 4k + 4)
k(k+1)^2 + 3k^2 + 7k + 4 = (k+1)(k+2)(k+2)
k(k+1)^2 + 3k^2 + 7k + 4 = (k+1)(k+2)^2

Therefore, the equation holds true for k+1 as well. By induction, the equation is proven to hold for all positive integers n.