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.
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