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.
 A: Hint: notice that $$\sum(2k)^2=\sum4k^2=4\sum k^2.$$
A: Let $$S=1^1+3^2+\cdots+(2n-1)^2=\sum_{k=1}^{n}(2k-1)^2=\sum_{k=1}^{n}[4 k^2-4k+1]$$
$$S=4\sum_{k=1}^n k^2- 4\sum_{k=1}^{n} k+\sum_{k=1}^{n} 1$$
$$S=4\frac{n(n+1)(2n+1)}{6}-4 \frac{n(n+1)}{2}+n=\frac{n(4n^2-1)}{3}$$
A: 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,
$$1^2+3^2+ \dots +(2n+1)^2=4\frac{n(n+1)(2n+1)}6+2(n+1)^2-1-n.$$
A: $ 1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$ multiplying this by $ 2^2 $ gives the sum $ 2^2+4^2+\dots + (2n)^2 = \frac{2n(n+1)(2n+1)}{3} $ and subtracting this from the given identity results int the required equality.
A: You have a typo. Call the known sum $P_n$. The desired sum is$$\begin{align}P_{2n+1}-4P_n&=\frac{(2n+1)(2n+2)(4n+3)-4n(n+1)(2n+1)}{6}\\&=\frac{(n+1)(2n+1)}{3}(4n+3-2n)\\&=\frac{(n+1)(2n+1)(2n+3)}{3}.\end{align}$$
A: Let $S=1^2+3^2+5^2+\cdots+(2n+1)^2$$.
Next, consider the sums
$$S_1=2^2+4^2+6^2+\cdots+(2n)^2$$
$$S_2=1^2+2^2+3^2+\cdots+(2n+1)^2$$
Thus, we can calculate $S$ as $S=S_2-S_1$.
Let's evaluate $S_1$ and $S_2$ using the identity we are given.
Clearly,
$$S_2=\frac{1}{6}(2n+1)(2n+2)(4n+3)$$
and
$$S_1=2^2(1^2+2^2+\cdots+n^2)$$
$$S_1=2^2\cdot\frac{1}{6}n(n+1)(2n+1)$$
Therefore,
$$S=S_2-S_1$$
$$S=\frac{1}{6}\left[2(n+1)(2n+1)(4n+3)-4n(n+1)(2n+1)\right]$$
$$S=\frac{(n+1)(2n+1)}{6}\left[8n+6-4n\right]$$
$$S=\frac{(n+1)(2n+1)(2n+3)}{3}$$
as desired.
Hope this helps :)
