Finding the sum of the series $1^2 + 2 × 2^2 + 3^2 + 2 × 4^2 + 5^2 + 2 × 6^2 + . . . + 2(n − 1)^2 + n^2 ,$ The following is a question which has been bugging me for quite a while, 


Find the sum of the series
$$1^2 + 2 × 2^2 + 3^2 + 2 × 4^2 + 5^2 + 2 × 6^2 + . . . + 2(n − 1)^2 + n^2 ,$$
Where $n$ is odd

I started by denoting this entire series with $S$, from there it is apparent that $S$ infact consists of two lesser series I shall call;

*

*$S_A$ The sum of the squares of odd natural numbers up till $n^2$

*$S_B$ Twice the sum of the squares of even natural numbers up till $(n-1)^2$
I resolved this would be easier to tackle by noting that
$$S_A + \frac{1}{2}S_B = \sum_{r=1}^{n} r^2$$
Which is just the sum of the squares of the first $n$ natural numbers
$$\therefore S_A +\frac{1}{2}S_B=\frac{n}{6}(n+1)(2n+1)$$
Leaving only the value of $\frac{1}{2}S_B$ to be found, this is where I am currently facing difficulty as I am unsure on whether my working is correct;


For the sum of the squares of the first n even natural numbers;
$$2^2 + 4^2 .... (2n)^2=2^2\sum_{r=1}^{n} r^2$$
$$\implies \frac{2}{3}n(n+1)(2n+1) $$
Hence the sum of the first $n-1$ even natural numbers should be
$$ \frac{2}{3}n(n-1)(2n-1)$$
And
$$S= \frac{n}{6}(n+1)(2n+1) + \frac{2}{3}n(n-1)(2n-1)$$
$$\therefore S= \frac{1}{6}n(10n^2 -9n + 5n) $$
However the correct answer is
$$\frac{1}{2}n^2(n+1)$$
Where has my working gone wrong and how would I arrive at the correct answer?

As implicitely pointed out in the given answer, your mistake is due to the fact that for the sum of the squares of even natural numbers up to $n-1$ we need to consider  the following:
$$2^2 + 4^2 .... (n-1)^2=2^2\left(1+2^2+\dots+\left(\frac{n-1}2\right)^2\right) $$
and not
$$2^2 + 4^2 .... (2n)^2=2^2\sum_{r=1}^{n} r^2$$
which is the sum of the squares of even natural numbers up to $2n$.
