Induction proof: $n^2+3n$ is even for every integer

Question

Prove using simple induction that $n^2+3n$ is even for each integer $n\ge 1$

I have made $P(n)=n^2+3n$ as the equation.

Checked for $n=1$ and got $P(1)=4$, so it proves that $P(1)$ is even.

Then I do it with random integer $k\ge 1$ and assume for P(k).

My issue is that I am not sure how to do $P(k+1)$, which is what I eventually need to prove the main equation.

Answer 1

$$P(k+1)=(k+1)^2+3(k+1)=k^2+2k+1+3k+3$$ Now can you express $P(k+1)$ in terms of $P(k)$? And once you've done that, what conclusion can you draw on $P(k+1)$, using the assumption that $P(k)$ is even?

Answer 2

$$P(k+1)=(k+1)^2+3(n +1)=\color{red}{k^2}+2k+1+\color{red}{3k}+3=\color{red}{P(k)}+2k+4$$

Answer 3

$$P(1)=4,$$ which we assume is known to be even. $$P(n+1)=(n+1)^2+3(n+1)=n^2+3n+2n+4=P(n)+2n+4.$$

Then, assuming that we can take for granted that $m$ and $m+2$ have the same parity,

$$P(n),P(n)+2,P(n)+4$$ and by induction $$P(n)+2n+4$$ are all even and so is $P(n+1)$.

Then by induction, all $P(n)$ are even.

Answer 4

I got it now. P(k+1)= P(k)+2k+4 can only give even values and hence it is proven P(n) can only have even answers.

Thank you everyone for the quick responses!