To prove the given formula using mathematical induction, we need to follow two steps:

Step 1: Verify the formula for the base case, n=1. Step 2: Assume the formula holds for some arbitrary value of n=k, and use it to prove the formula for n=k+1.

Let’s begin with Step 1:

Step 1: Base case

When n=1, the left-hand side of the formula becomes:

12 = 1(2×1 – 1)(2×1 + 1)/3 = 1×1×3/3 = 1

So the formula holds for the base case.

Step 2: Inductive step

Now we assume that the formula holds for some arbitrary value of n=k, and we want to prove that it holds for n=k+1. That is, we assume:

12 + 32 + 52 + … + (2k-1)² = k(2k-1)(2k+1)/3

And we want to prove:

12 + 32 + 52 + … + (2k-1)² + (2(k+1)-1)² = (k+1)(2(k+1)-1)(2(k+1)+1)/3

We can simplify the left-hand side of the above equation as follows:

12 + 32 + 52 + … + (2k-1)² + (2(k+1)-1)² = k(2k-1)(2k+1)/3 + (2k+1)²

= (2k+1)[k(2k-1)/3 + (2k+1)]

= (2k+1)[2k³ + k² – k³ + 6k² + 3k + 1]/3

= (2k+1)[2k³ + 7k² + 3k + 1]/3

= (2k+1)[2(k+1)³ – 3(k+1)² + (k+1)]/3

= (k+1)(2(k+1)-1)(2(k+1)+1)/3

Therefore, the formula holds for n=k+1 as well.

By the principle of mathematical induction, the formula holds for all positive integers n.

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