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

Posted January 10th, 2008 by Structure

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:

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.

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

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

Instant start