Answer
Let $k$, $k+1$, and $k+2$ be three arbitrary (but specific) consecutive integers. Then $k+(k+1)+(k+2)=3k+3=3(k+1)$. Clearly, $3(k+1)$ is divisible by $3$, since $k+1$ is an integer. Because the chosen consecutive integers were arbitrary, it must be that the sum of any three consecutive integers is divisible by $3$.
Work Step by Step
This is a direct proof by generalization from the generic particular, introduced in section 4.2. The proof hinges on the fact that the only information we know about $k$, $k+1$, and $k+2$ is that they are consecutive integers; hence, the proof is valid for all triples of consecutive integers.
