Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+3)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$ without induction? I've been trying to solve the following problem:

Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+1)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$.

I had the following idea, we write:
$$1^2+2^2+\dots + (2n+1)^2=\frac{(2n+1)((2n+1)+1)(2(2n+1)+1)}{6}=k$$
Let's pretend the identity we want to prove is true, then:
$$1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+1)}{3}=j$$
We take then $k-x=j$ and solve for $x$. If the given identities are true, $x$ must be the sum of $2^2+4^2+\dots+(2n)^2$, and we have that
$$x= \frac{2n (n+1) (2 n+1)}{3} $$
We still don't know that $x=2^2+4^2+\dots+(2n)^2$ but that can be easily proved by induction. I'd like to know: Is there some "neater" way that doesn't involve induction?
Despite the tag, I'd like to see an induction-free demonstration. I chose that tag because I couldn't think of anything better to choose.

Consider the identity $$(2k+1)^2=4k^2+4k+1$$ which relates the sum of the odd squares to the sum of the squares.
The extra terms $4k+1$ are easily dealt with with a telescoping,
$$(k+1)^2-k^2=2k+1.$$
Altogether,
$$1^2+3^2+ \dots +(2n+1)^2=4\frac{n(n+1)(2n+1)}6+2(n+1)^2-1-n.$$
