Answer
See below.
Work Step by Step
Proofs using mathematical induction consists of two steps:
1) The base case: here we prove that the statement holds for the first natural number.
2) The inductive step: assume the statement is true for some arbitrary natural number $n$ larger than the first natural number, then we prove that then the statement also holds for $n + 1$.
Hence here:
1) For $n=1: n^2-n+2=1^2-1+2=2$ and this is divisible by $2$.
2) Assume for $n=k: k^2-k+2$ is divisible by $2$. Then for $n=k+1:(k+1)^2-(k+1)+2=k^2+2k+1-k-1+2=k^2+k+2=k^2-k+2k+2$ and we know that $k^2-k$ is divisible by $2$ by the inductive hypothesis, $2k+2=2(k+1)$ is also divisible by $2$, thus their sum is also divisible by $2$. Thus we proved what we wanted to.