Using the principle of mathematical induction, prove each of the following for all n ϵ N:

n × ( n + 1 ) × ( n + 2 ) is multiple of 6


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): , which is multiple of 6


For n = 1 P(n) is true since , 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


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.


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