Proving $(2n+1) + (2n+3) + \cdots + (4n-1) = 3n^{2}$ for all positive integers $n$ 
Prove $(2n+1) + (2n+3) + (2n+5) +
 \cdots + (4n-1) = 3n^{2}$ for all
  positive integers $n$.

So the provided solution avoids induction and makes use of the fact that $1 + 3 + 5 + \cdots + (2n-1) = n^{2}$ however I cannot understand the first step: $(2n+1) + (2n+3) + (2n+5) + \cdots + (4n-1) = (1 + 3 + 5 + \cdots + (4n-1)) -(1 + 3 + 5 + \cdots + (2n-1))$. Once that has been established I can follow the rest, but I was hoping someone could help me understand why $(1 + 3 + 5 + \cdots + (4n-1))$ which already looks like less than the LHS can be made equal to the LHS by subtracting a positive number.
Additionally, I wanted to prove the equality using induction, but had trouble with that as well. I think I am thrown off by the last term $4n-1$. Even the initial case of $n=1$ is not totally clear to me: does it hold because $2(1)+1 = 3(1)^{2}$ or is it because $4(1)-1 = 3(1)^{2}$?
Either way, my approach was to replace every $n$ on the LHS with $n+1$ which resulted in: $$(2(n+1)+1) + (2(n+1)+3)+ \cdots + (4(n+1)-1)$$ and I am not sure what the second to last term in the sequence would be... I tried simplifying anyway $$(2n+3) + (2n+5) + \cdots + (4n+3)$$ and I this point I thought I could subtract $(2n+1)$ from both sides of the induction assumption resulting in $(2n+3) + (2n+5) + \cdots ? = 3n^{2} - (2n+1) - (4n-1)$ substituting this yields: $$3n^{2} - 2n - 1 - 4n + 1 + 4n + 3 = 3n^{2} + 6n + 3 - 8n + 1 = 3(n+1)^{2} - 8n + 1$$ but I guess I don't want the $-8n + 1$... I also tried substituting $3n^{2} - (2n+1)$ without that last term being subtracted, but that did not work out either. If anyone can help me understand how to do this properly I would really appreciate it. Thanks!

The simple solution using $1+3+\ldots+(2n-1)=n^2$ is as follows:
$(2n+1)+(2n+3)+\ldots+(4n-1)=2n\cdot n+ (1+3+\ldots+(2n-1))=3n^2$
The fact that $1+3+\ldots+(2n-1)=n^2$ is true is a simple induction: Assume it's true for $n$. Then for $n+1$ we get $1+3+\ldots+(2n-1)+(2(n+1)-1)=n^2+2n+1=(n+1)^2$
