Answer
Let $P(n)$ denote the proposition: "2 divides $n^2 +n$ whenever $n$ is a positive integer."
$P(1)$ is true, since $(1)^2 +(1) = 2$.
Assume that $P(k)$ is true for an arbitrary integer $k$. Then
\begin{align*}
(k+1)^2 + (k+1) &= k^2 + 3k +2 \\
&= [k^2 +k] + [2k] + [2]
\end{align*}
Each of the bracketed terms is divisible by $2$, so $P(k+1)$ is also true. Therefore by method of mathematical induction $P(n)$ is true - 2 divides $n^2 +n$ whenever $n$ is a positive integer.
Work Step by Step
First, we must give the proposition. We let $P(n)$ denote the proposition: "$2$ divides $n^2 +n$ whenever $n$ is a positive integer."
Next, we must prove our base case $P(1)$ is true by showing that the proposition is true when $n=1$. When $n=1$, $n^2 +n = (1)^2 +(1) =2$. $2$ is divisible by $2$, so therefore $P(1)$ is true.
Finally, we must carry out the inductive step, Assume that $P(k)$ is true; that is, we assume that $k^2 +k$ is divisible by $2$ for an arbitrary positive integer $k$. Now we must show that this implies that $P(k+1)$ is also true.
\begin{align*}
(k+1)^2 + (k+1) &= k^2 +2k +1 +k +1 \\
&= k^2 + 3k +2 \\
&= k^2 + k +2k +2 \\
&= [k^2 +k] + [2k] + [2]
\end{align*}
The first bracketed term is divisible by two, via our inductive hypothesis; the second and third bracketed terms are trivially divisible by two. This means that the entire expression must also be divisible by two, and that $P(k+1)$ is true. Therefore by method of mathematical induction $P(n)$ is true - 2 divides $n^2 +n$ whenever $n$ is a positive integer.