Find the sum of the first $n$ terms of the series $1^2+3^2+5^2+\ldots$.

The nth odd number is $2n-1$, so the nth term is $$t_n=(2n-1)^2=4n^2-4n+1.$$Therefore $$S_n=\sum_{k=1}^n (4k^2-4k+1)=4\sum_{k=1}^n k^2-4\sum_{k=1}^n k+n.$$Substituting the formulas for $\sum k^2$ and $\sum k$ and simplifying gives $$S_n=\frac{n}{3}(4n^2-1).$$Option D is the sum of cubes of the first $n$ odd natural numbers, not squares.