Prove the statement by the Principle of Mathematical Induction: $2+4+6+\ldots+2n = n^2+n$ for all natural numbers $n$.

(N/A) Let $P(n): 2+4+6+\ldots+2n = n^2+n$.
Step $1$: For $n=1$,$L.H.S. = 2$ and $R.H.S. = (1)^2+1 = 2$.
Since $L.H.S. = R.H.S.$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some natural number $k$,i.e.,$2+4+6+\ldots+2k = k^2+k$.
Step $3$: For $n=k+1$,we need to show $P(k+1): 2+4+6+\ldots+2k+2(k+1) = (k+1)^2+(k+1)$.
$L.H.S. = (2+4+6+\ldots+2k) + 2(k+1)$
$= (k^2+k) + 2k+2$
$= k^2+3k+2$
$= (k^2+2k+1) + (k+1)$
$= (k+1)^2 + (k+1) = R.H.S.$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
By the Principle of Mathematical Induction,$P(n)$ is true for all $n \in \mathbb{N}$.