=∑_(x=1)^n〖(3x)〗^2
S =∑_(x=1)^n (3x)^2 ( the sum, from x = 1 to x = n, of (3x)^2 )
S =9∑_(x=1)^n x^2
Now, ∑_(x=1)^n x^2 is the sum of the squares of the 1st n natural numbers and has a standard formula, S_n^2, where
S_n^2 = (n/6)(n + 1)(2n + 1)
Therefore, S = 9S_n^2
S = 1.5n(n + 1)(2n + 1)