Mistake in basic algebra, I think? Problem prove $(2n+1)+(2n+3)+(2n+5)+...+(4n-1) =3n^2$
Induction proof: base case $n=1$ assume true for all $n$ prove for $n+1$.
The $n$th or last term becomes $(4(n+1)-1)=4n+3$.
We also sub $n+1$ in for all $n$ the $n-1$ term is $(4n-1)$ and the first term is $2(n+1)+1=2n+3$
The right side is $3(n+1)^2 = 3(n^2 + 2n +1 ) $
Next thing is I appear to be missing the first term in the sum $(2n+1)$ on the left.
Adding $(2n+1)$ to both sides and subtracting $4n+3$ from both sides we get the $n$ case that equals $ 3n^2 $ on the left and
$$3n^2 + 6n +3  + 2n+1 -4n -3= 3n^2 + 4n+1$$
Which leaves me with $(2n+1)+(2n+3)+(2n+5)+...+(4n-1) =3n^2 +4n +1$
Which is approximately $4n+1$ on the right bigger than what I started with and is exactly the same on the left algebraically I must of done something wrong.
Which leads me to my question, what was it?

The left hand side is the sum of the odd numbers from $2n+1$ up to $4n-1$. Thus when going $n\to n+1$, the first summand $2n+1$ is dropped, and two summands $4n+1$ and $4n+3$ are appended a the end.
