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.
$\sum_{k=1}^{2k} (2k-1) + (2k+1) = 4k^2 + (2k+1)$ $\sum_{k=1}^{2k} (2k-1) + (2k+1) = 4k^2 + (2k+1)$
You wrote $P(n)=1+3+\cdots +2n$. This is not $P(n)$. For one, that last term is an even number, and you want to only sum odd numbers.
What you want is the first $2n$ odd numbers. The first odd number is $1=2(\color{red}{1})-1$. The second odd number is $3=2(\color{red}{2})-1$. The third odd number is $5=2(\color{red}{3})-1$. The fourth odd number is $7=2(\color{red}{4})-1$. I hope you see the pattern.
So $$P(n)=\sum_{k=1}^{2n} (2k-1) = 1+3+\cdots +4n-1$$