Through induction show that $\sum_{k=0}^n (2k+1) = (n+1)^2, n\in \mathbb{N}$

Question

Through induction I want to show that $$\sum_{k=0}^n (2k+1) = (n+1)^2, n\in \mathbb{N}$$ So I started of with $\sum_{k=0}^n (2k+1)=1+3+5+7+...+(2n+1)$ and started of with the base case, where $n=0$ $$(2(0)+1)=(0+1)^2$$ $$1=1$$ but the problem is that it doesn't hold for the next numbers, e.g. if $n=1$ then $$(2(1)+1)=(1+1)^2$$ $$3=4$$ which isn't true. What am I missing here because the original statement is correct.

Answer 1

The sum is from $k=0$ to $k=n$. So, you should have two terms:

$(2(0)+1) + (2(1)+1) = (1+1)^2$

is true.

Answer 2

$n^2+(2n+1)=(n+1)^2$. You should add ALL the odd numbers upto $2n+1$. $3$ is just the second odd number, the required sum for $n=1$ would be $1+3=4$ as required.