Prove n(n + 1) is Even by Mathematical Induction

Let P(n) denote the statement n(n + 1) is even.

Step 1 :

Substitute n = 1.

P(1) = 1(1 + 1)

P(1) = 1(2)

P(1) = 2, which is an even number

Therefore, P(1) is true.

Step 2 :

Substitute n = k.

P(k) = k(k + 1)

Let us assume that the statement P(n) be true for n = k

Then,

k(k + 1) is an even number ----(1)

Step 3 :

Substitute n = k + 1.

P(k + 1) = (k + 1)(k + 1 + 1)

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

Let y = k + 1 on the right side.

P(k + 1) = y(k + 2)

Use the distributive property on the right side.

P(k + 1) = ky + 2y

Replace y by (k + 1).

P(k + 1) = k(k + 1) + 2(k + 1) ----(2)

From (1), k(k + 1) is an even number.

2(k + 1) is an even number, as 2(k + 1) is a multiple of 2.

k(k + 1) + 2(k + 1) = an even number + an even number

(Sum of two even numbers is always an even number)

Then,

k(k + 1) + 2(k + 1) = an even number

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

P(k + 1) = an even number

So, P(k + 1) is true.

Thus, if P(k) is true, then P(k + 1) is also true.

Step 4 :

Therefore, by the principle of Mathematical induction, P(n) is true for all n ∈ N.

That is, n(n + 1) is even for all n ∈ N.

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