Prove that 1^2 +3^2 +5^2 +···+(2n+1)^2 =(n+1) (2n + 1)(2n + 3)/3 whenever n is a nonnegative integer

Let’s revisit the principle of mathematical induction.

PRINCIPLE OF MATHEMATICAL INDUCTION To prove that P(n) is true for all positive integers n, where P (n) is a propositional function, we complete two steps:

1. BASIS STEP: We verify that P (1) is true.
2. INDUCTIVE STEP: We show that the conditional statement P (k) → P (k + 1) is true for all positive integers k.

The definition above is from the textbook Discrete Mathematics and its Applications by Rosen.

Now, let’s solve the exercise.

As the principle states, we first need to do the basis step.

In this case, we need to verify that P(0) is true (instead of P(1)) because the theorem holds for all nonnegative integers and the first one is 0.

P(0): 1²=(0+1)(2×0+1)(2×0+3)/3

           1 = (1 x 1 x 3)/3 = 1

This completes the basis step.

In the inductive step, we first need to state P(k) and then prove that P(k+1) is true by using P(k).

P(k) = 1² +3² +5² +···+(2k+1)² =(k+1) (2k + 1)(2k + 3)/3

To state P(k+1), we do two things. On the left side of the equation, we add the term that comes after the k^th term (a.k.a. k+1). On the right side of the equation, we simply substitute k by k+1.

P(k+1) =1² +3² +5² +···+(2k+1)² + (2(k+1)+1)²=((k+1)+1) (2(k+1) + 1)(2(k+1) + 3)/3

Now, let’s simplify.

1² +3² +5² +···+(2k+1)² + (2k+3)²=((k+2) (2k+3)(2k+5)/3

Substituting P(k) in P(k+1). This is the inductive step. Here we are proving that if P(k) is true, then P(k+1) is also true.

(k+1) (2k + 1)(2k + 3)/3 + (2k+3)²=((k+2) (2k+3)(2k+5)/3

(k+1) (2k + 1)(2k + 3)/3 + 3(2k+3)²/3=((k+2) (2k+3)(2k+5)/3

[(k+1) (2k + 1)(2k + 3) + 3(2k+3)²]/3=((k+2) (2k+3)(2k+5)/3

[(k+1) (2k + 1)(2k + 3) + 3(2k+3)²]/3=((k+2) (2k+3)(2k+5)/3

(2k+3) [(k+1) (2k + 1) + 3(2k+3)]/3=((k+2) (2k+3)(2k+5)/3

(2k+3) [2k²+k+2k+1 + 6k+9]/3=((k+2) (2k+3)(2k+5)/3

(2k+3) [2k²+3k+1 + 6k+9]/3=((k+2) (2k+3)(2k+5)/3

(2k+3) [2k²+9k+10]/3=((k+2) (2k+3)(2k+5)/3

 (2k+3) [(k+2)(2k+5)]/3=((k+2) (2k+3)(2k+5)/3. ⧠

Notice that if you don’t remember how to do the factorization of 2k²+9k+10, you can always expand (k+2)(2k+5).

(k+2)(2k+5) = 2k²+5k+4k+10 = 2k²+9k+10.

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