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Question Number 158421 by tebohlouis last updated on 03/Nov/21

Answered by 1549442205PVT last updated on 04/Nov/21

In three consecutive integers there is always a number divisible by 3 and least at an integer to be even Hence (n−2)(n−1)n is divisible by 6 ,so (((n−2)(n−1)n)/6) is an integer

Inthreeconsecutiveintegersthereisalwaysanumberdivisibleby3andleastatanintegertobeevenHence(n−2)(n−1)nisdivisibleby6,so(n−2)(n−1)n6isaninteger

Answered by Rasheed.Sindhi last updated on 05/Nov/21

By Induction as required p(n)=((n(n−1)(n−2))/6) case-1: p(1)=0 (a whole number) case-2:Assuming p(k) is whole number. ⇒((k(k−1)(k−2))/6) is whole ⇒((k(k−1)(k−2))/6)+((k(k−1))/2) is whole [∵ One of k & k−1 is certainly even ∴ ((k(k−1))/2) is whole and whole+whole=whole] ⇒((k(k−1)(k−2)+3k(k−1))/6) is whole ⇒((k(k−1)(k−2+3))/6) is whole ⇒((k+1^(−) (k+1^(−) −1)(k+1^(−) −2))/6) is whole ⇒p(k+1) is whole when p(k) is whole ∴ p(n) is true for all natura numbers.

ByInductionasrequiredp(n)=n(n−1)(n−2)6case−1:p(1)=0(awholenumber)case−2:Assumingp(k)iswholenumber.⇒k(k−1)(k−2)6iswhole⇒k(k−1)(k−2)6+k(k−1)2iswhole[∵Oneofk&k−1iscertainlyeven∴k(k−1)2iswholeandwhole+whole=whole]⇒k(k−1)(k−2)+3k(k−1)6iswhole⇒k(k−1)(k−2+3)6iswhole⇒k+1―(k+1―−1)(k+1―−2)6iswhole⇒p(k+1)iswholewhenp(k)iswhole∴p(n)istrueforallnaturanumbers.

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