Prove that for all integers $n\geq1$, $P(n) = 2+4+6+...+2n=n^2+n$
For $P(1)$, we have that: $2\cdot1=1^2+1$
Let $k$ be an integer with $k\geq1$. Suppose $P(k)$ is true. Then: $2+4+6+...+2k=k^2+k$
I must show that $P(k+1)$ is also true:
$2+4+6+...+2(k+1)=(k+1)^2+(k+1)$ $(1)$
If $k$ is an integer then $l=k+1$ is also an integer. For $l=k+1$ $(1)$ becomes: $l^2+l$ which is of the form $k^2+k$ that is true.
$Q.E.D$
A small change at the inductive step:
I must show that $P(k+1)$ is also true:
$(2+4+6+...+2k)+2(k+1)=k^2+k+2k+2=(k^2+2k+1)+(k+1)=(k+1)^2+(k+1)$
$Q.E.D$
I know that the second inductive step is definitely correct, but I'm confused as to whether the first one is correct.