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We need to show that, based on that assumption, (k+1)(k+2)(k+3) is also
divisible by 6.
(k+1)(k+2)(k+3) = (k+1)(k+2)k + (k+1)(k+2)3 = k(k+1)(k+2) + 3(k+1)(k+2).
By induction hypothesis, the first term is divisible by 6,
and the second term 3(k+1)(k+2) is divisible by 6 because it contains
a factor 3 and one of the two consecutive integers k+1 or k+2 is
even and thus is divisible by 2. Thus it is divisible by both 3 and
2, which means it is divisible by 6. The theorem is proved since
the sum of two multiples of 6 is also a multiple of 6.

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

Let the numbers be n,n+1,n+2 For a number to be divisible by 6, it must be divisible by 2 and 3 both.

Out of n,n+1,n+2 , one number will be divisible by 2 because if any one of the numbers is odd ,the next number will be even and hence divisible by 2.

Similarly we can say that since the multiples of 3 repeat after intervals of 3, one in 3 consecutive numbers will be divisible by 3.