For any positive integer, prove that divisible by 6.

We have,

Given that (S) = n3- n is divisible by 6.


Let n =1 then we get ‘0’ which is divisible by 6.


Therefore S(1) is true.


Let’s take n = p Therefore, S(p) = p3 - p which is divisible by 6.


= t (an integer)


(p3 - p) = 6t


p3 = 6t + p ----- Eqn (1)


Now take n = p + 1


(p+1)3 - (p+1) (p3+3p2+3p+1) - (p+1) --- Eqn (2)


Substituting valve of Eqn (1) in Eqn 2, we get


(6t+p)3 + 3p2 +2p – (p+1)


6t +3p(p+1)


Let p(p+1) = 2a is an even number a is natural number


Therefore 6t+3×2a => 6t+6a => 6(t+a) which is divisible by 6


Therefore for any positive integer of form divisible by 6.


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