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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