Proving that $n^2 + n$ is even for any integer $n$ This type of question often shows up in introductory abstract math courses, computer science discrete math, or an introduction to logic course. 
Show that $n^2 + n$ is even for any integer n.
 A: $n^2+n=n(n+1)$ and either $n$ is even or $n+1$ is even. In any case, their product is even.
A: $$n^2+n=n(n+1)$$
So the above is the product of two consecutive integers and then one of them exactly must be even and thus so is the product.
A: Before beginning, you should understand the following:
even * even = even
even + even = even
odd  + odd  = even
odd  * odd  = odd

First, suppose $n$ is odd:
n = 2k + 1 is the formula of an odd number $n$ (this means $2k+1$ always produces an odd number $n$, try it by plugging in any integer for $k$).
Let's use this fact of n = 2k + 1 with the expression we are trying to prove is always even; remember the original expression? It is: $$n^2 + n$$ is always even.
Second, plug in for $n$:
Using our odd number formula that we know n = 2k + 1 
and our original expression $n^2 + n$, plug in
$n^2 + n = (2k+1)^2 + (2k+1)$
Using the right-hand side:
Expand it by "foiling":
$(2k+1)^2 + (2k+1)$ 
$(4k^2 + 4k + 1) + (2k + 1)$ 
Simplified: $4k^2 + 6k + 2$
Bring back the left-hand side and put it all together:
$n^2 + n = 4k^2 + 6k +2$
Analyzing the right-hand side you can see that no matter what $k$ is, it is multiplied by 4, which makes it always even (this applies to both the $k$'s. Same goes for the $6k$ because 6 is even. And as it always has been: 2 will always be an even number.
Now you need to suppose $n$ is even, instead of odd
Do the same steps you did above but instead of n = 2k + 1 for all odd numbers, you need to know that n = 2k for all even numbers. 
Take your original expression that you're trying to prove is always even: $n^2 + n$ and your newly discovered even number formula $n = 2k$ then plug in just like you did above.
A: Here is a different take. Use Gauss's trick:
$S = 1 + 2 + \cdots + n$
$S = n +(n-1) + \cdots + 1$
$2S = (n+1) + (n+1) + \cdots + (n+1) = n(n+1)$
Thus, $n(n+1)=2S$ is even.
