Sum of $1^2+3^2+\cdots+(2n+1)^2$ Have trouble with proof. I've been working through a question on Courant's What is Mathematics?
This is the question: Prove $1^2+3^2+\cdots+(2n+1)^2=\frac{n(n+1)(2n+1)(2n+3)}{3}$. I called this $S_{(2n+1)^2}$.
What I've been given as a hint is that I can use the sum $1+2^2+3^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ (I called this $S_{n^2}$) in some way.
So here's my line of thought:
What do I need to do to $S_{n^2}=1^2+2^2+3^2+\cdots+n^2$ to get $S_{(2n+1)^2}=1^2+3^2+\cdots+(2n+1)^2$?
When I compared them, I thought that $S_{n^2}-(2^2+4^2+\cdots+(2n)^2) = S_{(2n+1)^2}$.
I defined $2^2+4^2+\cdots+(2n)^2$ as $S_{(2n)^2}$.
After a bit of head-scratching I managed to figure out $S_{(2n)^2}=2^2(0)+2^2(1)^2+2^2(2)^2+\cdots+2^2(n)^2=4(1^2+2^2+3^2+\cdots+n^2)=4S_{n^2} =\frac{2}{3}n(n+1)(2n+1)$
Now is where my confusion starts. So from my logic above I went forward with $S_{n^2}-S_{(2n)^2}=S_{(2n+1)^2}$ which gave me $-\frac{1}{2}(n(n+1)(2n+1))$ which isn't $S_{(2n+1)^2}$.
I don't know where I really went wrong. One of my main suspicions is that $S_{n^2}-S_{(2n)^2}=S_{n^2}-4S_{n^2}=-3S_{n^2}\neq S_{(2n+1)^2}$. But I don't see why.
 A: Very easy proof
$$\sum_{k=1}^n(2k+1)^2=4\sum_{k=1}^n k^2+4\sum_{k=1}^nk+\sum_{k=1}^n1.$$
Using your hint, you can very easily conclude.
A: $$1^2+2^2+\cdot\cdot\cdot+n^2=\dfrac{n(n+1)(2n+1)}{6}$$
This leads to the following:

$$1^2+\cdot\cdot\cdot+n^2+\cdot\cdot\cdot+(2n+1)^2=\dfrac{(2n+1)(2n+1+1)(2(2n+1)+1)}{6}$$
$$2^2+4^2+\cdot\cdot\cdot+(2n)^2=2^2(1^2+2^2+\cdot\cdot\cdot+n^2)=4\cdot\dfrac{n(n+1)(2n+1)}{6}$$

Thus,

$$1^2+3^2+\cdot\cdot\cdot+(2n+1)^2=\dfrac{(2n+1)(2n+1+1)(2(2n+1)+1)}{6}-\quad4\cdot\dfrac{n(n+1)(2n+1)}{6}=\dfrac{(n+1)(2n+1)(2n+3)}{3}=\sum_{i=0}^n (2i+1)^2$$

Note: You have an extra $n$ in your answer.
A: You are using a misleading notation. Let
$$
Q_n=\sum_{k=1}^n k^2=1^2+2^2+\dots+n^2={n(n+1)(2n+1)\over6},
$$
so that
$$
Q_{2n+1}=\sum_{k=1}^{2n+1} k^2=1^2+2^2+\dots+(2n)^2+(2n+1)^2={(2n+1)(2n+2)(4n+3)\over6}.
$$
As you noticed:
$$
1^2+3^2+\dots+(2n-1)^2+(2n+1)^2=Q_{2n+1}-4Q_n={(n+1)(2n+1)(2n+3)\over3}.
$$
A: Hint: Let $S_k$ denote the sum of the first $k$ squares. You should know a formula for this. Now
$$1^2+3^2+\cdots+(2n+1)^2$$
$$= (1^2+2^2+\cdots+(2n+2)^2)-(2^2+4^2+\cdots+(2n+2)^2)$$
$$=(1^2+2^2+\cdots+(2n+2)^2)-2^2(1^2+2^2+\cdots+(n+1)^2)$$
$$=S_{2n+2}-4S_{n+1}$$
Can you take it from here?
A: Using your hint, here is one way to go:
$$1^2 + 3^2 +\cdots +(2n+1)^2 = (1^2 + 2^2 + 3^2 + 4^2+\cdots +(2n+1)^2) - (2^2 + 4^2 + 6^2+\cdots+ (2n)^2)\\
(1^2 + 2^2 + 3^2 + 4^2+\cdots+ (2n+1)^2) - 4(1^2 + 2^2 + 3^2+\cdots+ n^2)$$
$$\frac{(2n+1)(2n+2)(4n+3)}{6} - \frac{4(n)(n+1)(2n+1)}{6}$$
Alternatively, you could use a proof by induction.
Base case:
$$n = 1\\
1^2 + 3^2 = \frac {1\cdot2\cdot3\cdot5}{3} = 10$$
Suppose:
$$1^2+3^2+\cdots+(2n+1)^2=\frac{n(n+1)(2n+1)(2n+3)}{3}$$
We will show that:
$$1^2+3^2+\cdots+(2n+1)^2 + (2n+3)^2=\frac{(n+1)(n+2)(2n+3)(2n+5)}{3}$$
By our hypothesis:
$$\frac{n(n+1)(2n+1)(2n+3)}{3} + (2n+3)^2=\frac{(n+1)(n+2)(2n+3)(2n+5)}{3}$$
And simplify.
