After finding the formula in a), we will need to use induction to prove that it holds for the first n even positive numbers. So, let’s refresh the principle of mathematical induction.

PRINCIPLE OF MATHEMATICAL INDUCTION To prove that P(n) is true for all positive integers n, where P (n) is a propositional function, we complete two steps:

BASIS STEP: We verify that P (1) is true.
INDUCTIVE STEP: We show that the conditional statement P (k) → P (k + 1) is true for all positive integers k.

The definition above is from the textbook Discrete Mathematics and its Applications by Rosen.

Now, let’s solve the exercise.

a) Find a formula for the sum of the first n even positive integers

We start by writing the sum and the result, so we can find the general term.

(n=1) 2 = 2

(n=2) 2 + 4 = 6

(n=3) 2 + 4 + 6 = 12

(n=4) 2 + 4 + 6 + 8 = 20

(n=5) 2 + 4 + 6 + 8 + 10 = 30

(n=6) 2 + 4 + 6 + 8 + 10 + 12 = 42

The general term will be 2 + 4 + 6 + … + (2n-4) + (2n-2) + 2n

Let’s re-write it by grouping 2 terms as follows:

(2 + 2n) + (4 +2n-2) + (6 + 2n-4) + … = (2+2n) + (2n + 2) + (2n + 2) + …

Notice on the right side of the equation, that because we arrange the summation in pairs, and there are in total n terms, the number of pairs is n/2.

(2+2n) + (2n + 2) + (2n + 2) + … = n/2(2n+2)

= (2n²+2n)/2

=2(n²+n)/2

= n²+n

= n(n+1)

You can check if this term gives the right results for n=1 to n=6.

Now let’s prove that it holds for every n.

b) Prove the formula that you conjectured in part (a)

As the mathematical induction principle states, we first need to do the basis step.

Basis step:

P(1): 2 = 1(1+1) = 1×2 = 2

Inductive step:

P(k): 2 + 4 + 6 + … + 2k = k(k+1)

P(k+1): 2 + 4 + 6 + … + 2k + 2(k+1) = (k+1)(k+1+1)

2 + 4 + 6 + … + 2k + 2(k+1) = (k+1)(k+2)

Substituting P(k) in P(k+1) we get,

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

Common factor,

(k+1)( k + 2) = (k+1)(k+2) ⧠

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