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 $n=1: 1^2+1=2$ and this is divisible by $2$.
2) Assume for $n=k: k^2+k$ is divisible by $2$. Then for $n=k+1:(k+1)^2+(k+1)=k^2+2k+1+k+1=k^2+3k+2$ 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.
