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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:

1. Base case (n=1): Show that the statement is true for the lowest value of n.

2. 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.

3. 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.

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 + 2k, where m and k are integers.

We can factor out a 2 from the right-hand side:

2(m + k)

Since m and k are integers, m+k is also an integer. Let's say m+k=n, where n is an integer. Thus, we have:

2n

This expression shows that k^2 + k is divisible by 2. Therefore, the statement holds for n=k+1.

By the principle of mathematical induction, we've shown that n^2 - n is always divisible by 2 for all positive integers n.