prove with inductions that n^2+n divisible by 2

To show that $n^2+n$ is divisible by 2, we must show that $n^2+n=2M$ for some $M \in \mathbb{Z}$ by induction.

For $n=1$, we have $1^2+1=2(1).$ Here, $M=1\in \mathbb{Z}$ . So, it is true for $n=1$ .

Assume it is true for $n=k, k\in \mathbb{Z}^+$ . That is,

$k^2+k=2M .....................................(1)$

For $n=k+1$ . We have;

$(k+1)^2+(k+1)=k^2+2k+1+k+1\\ =k^2+3k+2....................................(2)$

From (1), $k^2=2M-k.$ Put this into (2)

$=2M-k+3k+2\\ =2M+2k+2\\ =2(M+k+1)\\ =2M'$ where $M'=(M+k+1)\in\mathbb{Z}$ .

Therefore, it is true for $n=k+1$ .

Hence, $n^2+n$ is divisible by 2.