Prove $n^{2}+n$ is always even.

Step 1 ：Given the expression \(n^{2}+n\), we can rewrite this as \(n(n+1)\).

Step 2 ：For any integer n, either n or n+1 is always even. Therefore, the product of n and n+1, i.e., \(n^{2}+n\) is always even.

Step 3 ：We can also prove this by taking two cases - when n is even and when n is odd.

Step 4 ：Case 1: When n is even, we can write it as 2k (where k is an integer). Substituting this in the given expression, we get \(4k^{2}+2k = 2(2k^{2}+k)\), which is clearly even.

Step 5 ：Case 2: When n is odd, we can write it as 2k+1. Substituting this in the given expression, we get \((4k^{2}+4k+1)+(2k+1) = 2(2k^{2}+3k+1)\), which is also even.

Step 6 ：Therefore, in both cases, \(n^{2}+n\) is always even.

Step 7 ：\(\boxed{\text{The final answer is that the expression } n^{2}+n \text{ is always even for any integer n.}}\)