To Prove:
n × ( n + 1 ) × ( n + 2 ) is multiple of 6
Let us prove this question by principle of mathematical induction (PMI) for all natural numbers
n × ( n + 1 ) × ( n + 2 ) is multiple of 6
Let P(n): n x (n + 1) x (n + 2), which is multiple of 6
For n = 1 P(n) is true since 1 x (1 + 1) x (1 + 2) = 6, which is multiple of 6
Assume P(k) is true for some positive integer k , ie,
= k × (k + 1) × (k + 2) = 6m, where m ∈ N …(1)
We will now prove that P(k + 1) is true whenever P( k ) is true
Consider,
= (k + 1) × ((k + 1) + 1) × ((k + 1) + 2)
= (k + 1)×{ k + 2 }×{ (k + 2) + 1}
= [(k + 1)×(k + 2)× (k + 2) ] + (k + 1)×(k + 2)
= [ k×(k + 1)×(k + 2) + 2×(k + 1)×(k + 2) ] + (k + 1)×(k + 2)
= [6m + 2×(k + 1)×(k + 2) ] + (k + 1)×(k + 2)
= 6m + 3×(k + 1)×(k + 2)
Now, (k + 1) & (k + 2) are consecutive integers, so their product is even
Then, (k + 1) × (k + 2) = 2×w (even)
Therefore,
= 6m + 3 × [2×w ]
= 6m + 6×w
= 6(m + w)
= 6×q where q = (m + w) is some natural number
Therefore
(k + 1) x ((k + 1) + 1) x ((k + 1) + 2) is multiple of 6
Therefore, P (k + 1) is true whenever P(k) is true.
By the principle of mathematical induction, P(n) is true for all natural numbers, ie, N
Hence proved.
