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 $n=1: 1^2-1+41=41$ is odd.
2) Assume for $n=k: k^2-k+41$ is odd. Then for $n=k+1$:
$(k+1)^2-(k+1)+41=k^2+2k+1-k-1+41=k^2+k+41=(k^2-k+41)+2k$
The first part of the sum is odd by the induction hypothesis, whilst the second part is divisible by $2$ because of the factor. The sum of an odd and an even number is odd, hence the sum is also odd.
Thus we proved what we wanted to.