Precalculus, Chapter 9, 9.4, Section 9.4, Problem 19

You need to use mathematical induction to prove the formula for every positive integer n, hence, you need to perform the two steps of the method, such that:
Step 1: Basis: Show that the statement P(n) hold for n = 1, such that:
1 = 1/3*(2*1-1)(2*1+1) => 1 =3/3 => 1=1
Step 2: Inductive step: Show that if P(k) holds, then also P(k + 1) holds:
P(k): 1^2 + 3^2 + 5^2 + .. + (2k-1)^2 = (k(2k-1)(2k+1))/3 holds
P(k+1): 1^2 + 3^2 + 5^2 + .. + (2k-1)^2 + (2k+1)^2 = ((k+1)(2k+1)(2k+3))/3
You need to use induction hypothesis that P(k) holds, hence, you need to re-write the left side, such that:
(k(2k-1)(2k+1))/3 + (2k+1)^2= ((k+1)(2k+1)(2k+3))/3
k(2k-1)(2k+1) + 3(2k+1)^2= (k+1)(2k+1)(2k+3)
You need to factor out 2k+1:
(2k+1)(2k^2 - k + 6k + 3) = (2k+1)(2k^2 + 5k + 3)
(2k+1)(2k^2 + 5k + 3) = (2k+1)(2k^2 + 5k + 3)
Notice that P(k+1) holds.
Hence, since both the basis and the inductive step have been verified, by mathematical induction, the statement P(n): 1^2 + 3^2 + 5^2 + .. + (2n-1)^2 = (n(2n-1)(2n+1))/3 holds for all positive integers n.