Use mathematical induction to show that n^2-n is always divisible by 2. To prove that n^2 - n is always divisible by 2 using mathematical induction, we'll follow these steps: Base case (n=1): Show that the statement is true for the lowest value of n. Inductive step: Assume that the statement is true for n=k (k is an arbitrary positive integer), and show that it is also true for n=k+1. Base case (n=1): When n=1, the expression becomes: 1^2 - 1 = 0 Since 0 is divisible by 2, the statement holds for n=1. Inductive step: Assume that the statement is true for n=k, i.e., k^2 - k is divisible by 2. We can write: k^2 - k = 2m, where m is an integer. Now, we need to prove that the statement is true for n=k+1. So, let's consider the expression for (k+1)^2 - (k+1): (k+1)^2 - (k+1) = k^2 + 2k + 1 - k - 1 = k^2 + k Now, we want to show that k^2 + k is divisible by 2. To do this, we can rewrite the expression using the assumption that k^2 - k is divisible by 2: k^2 + k = (k^2 - k) + 2k = 2m ...