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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