Answer
See below.
Work Step by Step
Proofs using mathematical induction consist 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 we have:
1) For $n=1: 1^2+1=2$ is divisible by $2$.
2) Assume for $n=k: k^2+k$ is divisible by $2$. Then for $n=k+1$:
$(k+1)^2+k+1=k^2+2k+1+k+1=(k^2+k)+2(k+1)$
The first part of the sum is divisible by $2$ by the induction hypothesis, whilst the second part is divisible because of the factor. Hence, the sum is also divisible.
Thus we proved what we wanted to.