Question 1045633
i'm not real good at these, but let me give it a shot.


you want to prove that n * (n + 1) is even, when you are dealing with a natural number.


a natural number is all the positive integers i believe.


it does not include 0 as far as i remember.


you have 2 possibilities.


if n is even, then n + 1 is odd.


if no is odd, then n + 1 is even.


assuming that n is even, then there exists a number k such that 2 * k = n.


n * (n + 1) becomes 2 * k * (2 * k + 1) which becomes 4 * k^2 + 2 * k.


the number 4 * k^2 + 2 * k has to be even because it's divisible by 2.


assuming that n is odd, then there exists a number k such that 2 * k = n + 1.


n * (n + 1) becomes n * 2 * k which becomes 2 * k * n.


the number 2 * k * n  has to be even because it's divisible by 2.


any number that's divisible by 2 is even.


not sure if this constitutes an acceptable proof by your instructor, but it seems reasonable to me.