Proof by Induction $n^2+n$ is even

I'm not entirely sure if I'm going about proving $n^2+n$ is even for all the natural numbers correctly.

$P(n): = n^2+n$

$P(1) = 1^2+1 = 2 = 0$ (mod $2$), true for $P(1)$

Inductive step for $P(n+1)$:

$\begin{align}P(n+1) &=& (n+1)^2+(n+1)\\ &=&n^2+2n+1+n+1\\ &=&n^2+n+2(n+1)\end{align}$

Does this prove $n^2+n$ is even as it's divisible by $2$? Thanks!

Solution 1:

I see other answers provide different (possibly simpler) proofs. To finish off your proof:

by the induction hypothesis $n^2+n$ is even. Hence $n^2 + n = 2k$ for some integer $k.$ We have $$n^2+n+2(n+1) = 2k + 2(n+1) = 2(k+n+1) = 2\times\text{an integer} = \text{even}.$$

Does this prove $n^2+n$ is even as it's divisible by $2$?

The key here is to remember stating & using the induction hypothesis.

Solution 2:

Or this....

$$n^2 + n = n(n+1).$$ One of $n$ or $n+1$ is even so the product is even.