Use induction to show that $\displaystyle 2|n(n+1)$
$\displaystyle n=1, \ 2|2$
Assume $\displaystyle p(k)$ is true.
$\displaystyle 2|(k+1)(k+2)\Rightarrow 2|(k^2+3k+2)$
$\displaystyle 2m=k(k+1)$
Now what?
Assume $\displaystyle n = k$ is true.
$\displaystyle \Rightarrow 2|(k)(k+1) \Rightarrow k(k + 1) = 2m$
Show true for n = k + 1:
$\displaystyle \begin{aligned}
(k + 1)(k + 2) &= k(k + 1) + 2(k + 1) \\
&= 2m + 2(k + 1) \\
&= 2(m + k + 1) \\
&\Rightarrow 2|(k + 1)(k + 2)
\end{aligned}$
K or k+1 is even.
Any amount of an even value is even.
Via induction...
P(k)
k(k+1) is even ?
P(k+1)
(k+1)(k+2) is even ?
Try to prove that P(k) being true also causes P(k+1) to be true
Proof
$\displaystyle (k+2)(k+1)=k(k+1)+2(k+1)$
If P(k) is true, then P(k+1) is true as 2(k+1) is a multiple of 2 and so is even.