prove that $n(n+1)$ is even using induction Prove that $n(n+1)$ is even using induction


The base case of $n=1$ gives us $2$ which is even.

Assuming $n=k$ is true, 

$n=(k+1)$ gives us $ k^2 +2k +k +2$ 

while $k(k+1) + (k+1)$ gives us $k^2+2k+1.$

whats is the next step to prove this by induction? I can't seem to show 

$ k^2 +2k +k +2$ = $k^2+2k+1$
Hint:
$$(n+1)((n+1)+1)=(n+1)(n+2)=n(n+1)+2(n+1)$$
$n(n+1)$ is an even number.
Take any $n\in\mathbb N$, then $n$ is either even or odd.

Suppose $n$ is even $\implies n= 2m$ for some $m\implies n(n+ 1) = 2m(n+ 1)\implies n(n+ 1)$ is even.
Suppose $n$ is odd $\implies n+1$ is even $\implies n+1 = 2m$ for some $m \implies n(n+1) = 2nm\implies n(n+1)$ is even.