The sum of the squares of the first n odd natural numbers

What is the sum of the squares of the first n odd natural numbers?

The answer is:

$$\sum_{k=1}^{k=n} (2k-1)^2=1^2+3^2+5^2+...+(2n-1)^2=\frac{n(2n-1)(2n+1)}{3}$$

Proof 1:

We will use what we know from this:
The sum of the squares of the first n natural numbers

Let

$$S_{n,2}=\sum_{k=1}^{k=n}k^2=1^2+2^2+3^2+....+(n-1)^2+n^2=\frac{n(n+1)(2n+1)}{6}$$

We have

$$S_{2n,2}=\sum_{k=1}^{k=2n}k^2=1^2+2^2+3^2+....+(2n-1)^2+(2n)^2=\frac{2n(2n+1)(4n+1)}{6}$$

We want to find

$$S_{odd,n,2}=\sum_{k=1}^{k=n}(2k-1)^2=1^2+3^2+5^2+...+(2n-1)^2=$$
$$1^2+2^2+3^2+....+(2n-1)^2+(2n)^2-(2^2+4^2+6^2+...+(2n)^2)=S_{2n,2}-4S_{n,2}=$$
$$\frac{2n(2n+1)(4n+1)}{6}-4\frac{n(n+1)(2n+1)}{6}=\frac{2n(2n+1)(2n-1)}{6}=\frac{n(2n-1)(2n+1)}{3}$$

Proof 2 - Induction proof

1) For n = 1 it's true.

2) Let's consider that is true for n, and let's prove it for n+1.

$$\frac{n(2n-1)(2n+1)}{3}+(2n+1)^2=\frac{(2n+1)(2n^2-n+6n+3)}{3}=$$
$$=\frac{(2n-1)(2n^2+5n+3)}{3}=\frac{(n+1)(2n+1)(2n+3)}{3}$$
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