Proof by induction that the sum of the first $2n$ odd positive integers is $4n^2$

Prove by induction, explaining each step carefully, that the sum of the first $2n$ odd positive integers is equal to $4n^2$.

Let P(n) be the statement $P(n)=\sum_{n=1}^{2n} 2n-1 = 4n^2$

The $P(1)$ asserts that $(2(1)-1)+(2(2)-1)=1+3=4$, and we see that P(1) is T, so this establishes the basis for induction.

To verify the induction step, we suppose the P(k) is T, where $k\in \!\,\mathbb{N} \!\,$. That is, we assume:
$\sum_{k=1}^{2k} 2k-1 = 4k^2$

(now this is where I am screwing up)

Since we wish to conclude that P(k+1) is T, we add $2k+1$ to both sides.

$\sum_{k=1}^{2k} (2k-1) + (2k+1) = 4k^2 + (2k+1)$
$\sum_{k=1}^{2k} (2k-1) + (2k+1) = 4k^2 + (2k+1)$