Answer
Conditions $I$ and $II$ of the principle of mathematical induction both hold therefore
$2+4+6+...+2n=n(n+1)$ ifor all natural numbers $n$.
Refer to the step-by-step part below for the proof.
Work Step by Step
We are searching for the sum of even numbers.
For $n=1$, the given statement true, as
$2=1(1+1)$
Thus, $\bf{\text{Condition I}}$ holds.
Let us assume that the given satement is true for some natural number $k$, that is,
$$2+4+6+...+2k=k(k+1)$$
We need to show that the given statement holds for $k+1$, which is
$$2+4+6+...+2k + 2(k+1)=(k+1)[(k+1)+1]=(k+1)(k+2)$$
Using $2+4+6+...+2k=k(k+1)$, we have
\begin{align*}
2+4+6+...+2k + 2(k+1)&=(k+1)(k+2)\\
\left(2+4+6+...+2k\right) + 2(k+1)&=(k+1)(k+2)\\
k(k+1)+2(k+1)&=(k+1)(k+2)\\
(k+1)(k+2)&=(k+1)(k+2) &\text{(factored out (k+1) on the left side)}
\end{align*}
$\bf{\text{Condition II }}$ holds.
Thus, the given statement is true for all natural numbers $n$.
