I've been trying to find a formula for sum of the first n squares of odd numbers (a question from Spivak's calculus) and I was trying to subtract the the $i^{th}$ odd number from the $i^{th}$ even number, I've noticed that the the difference between the squares is (or seems to be) increasing by 4. This what I'm seeing: \begin{align*} 2 ^ 2 - 1 ^ 2 = 3\\ 4 ^ 2 - 3 ^ 2 = 7 \\ 6 ^ 2 - 5 ^ 2 = 11 \\ 8 ^ 2 - 7 ^ 2 = 15 \\ 10 ^ 2 - 9 ^ 2 = 19 \\ 12 ^ 2 - 11 ^ 2 = 23 \end{align*} Every consecutive term in $3,7,11,15,19,23$ differs by 4. The fact that the difference of the squares for the $i^{th}$ even and odd integers increases by 4 seems completely arbitrary. Why 4? Is there something going on that I'm just missing? I don't know if it has something to do with Dirichlet's arithmetic progression for $4n + 3$. The pattern indicates this formula to me: $$ \sum_{k=1}^{n} (2k)^2 = \sum_{k=1}^{n}(2k-1)^2 + \sum_{k=0}^{n-1}(4k+3) $$ I believe this formula is true (though I haven't proved it yet), this will make it possible for me to use the closed form: $$\sum_{k=1}^{2n}k^2 = \frac{n(2n+1)(4n+1)}{3}$$
to find the closed form for the sum of odd numbers, But the question is: why is the difference between the $n^{th}$ even and odd integer equal to $4(n-1)+3$?
You are asking about $$ (2n)^2 - (2n-1)^2 - \left((2n-2)^2-(2n-3)^2 \right) $$ which is always 4 by algebra.