By using mathematical induction, prove that the equation   2+4+6+...+2n = n(n+1) is true, for all positive integers. 

Solution:

To prove the equation 2 + 4 + 6 + ... + 2n = n(n + 1) using mathematical induction, we will follow the steps of the proof:

Step 1: Base Case

First, we need to verify that the equation holds true for the smallest possible value of n, which is 1.

For n = 1:

2(1) = 1(1 + 1) -> 2 = 2

Since the equation holds true for n = 1, the base case is satisfied.

Step 2: Inductive Hypothesis

Assume that the equation holds true for some arbitrary positive integer k, where k ≥ 1. This assumption is called the inductive hypothesis.

Assumption: 2 + 4 + 6 + ... + 2k = k(k + 1)

Step 3: Inductive Step

We need to prove that the equation holds true for the next positive integer, which is k + 1.

Consider the sum 2 + 4 + 6 + ... + 2k + 2(k + 1). This can be written as the sum of the first k terms plus the (k + 1)th term.

2 + 4 + 6 + ... + 2k + 2(k + 1) = [2 + 4 + 6 + ... + 2k] + 2(k + 1)

Using the inductive hypothesis, we can substitute k(k + 1) for the sum of the first k terms:

= [k(k + 1)] + 2(k + 1)

Now, let's simplify the right-hand side of the equation:

= k(k + 1) + 2(k + 1)

= k^2 + k + 2k + 2

= k^2 + 3k + 2

= (k + 1)(k + 2)

Therefore, we have shown that if the equation holds true for some positive integer k, then it also holds true for the next positive integer, k + 1.

Step 4: Conclusion

Since the base case (n = 1) holds true and we have shown that if the equation holds true for any positive integer k, it also holds true for k + 1, we can conclude that the equation 2 + 4 + 6 + ... + 2n = n(n + 1) is true for all positive integers.