Answer :

Given; P(n) = n(n2 + 5) is divisible by 6.

P(0) = 0(02 + 5) = 0 ; is divisible by 6.


P(1) = 1(12 + 5) = 6 ; is divisible by 6.


P(2) = 2(22 + 5) = 18 ; is divisible by 6.


P(3) = 3(32 + 5) = 42 ; is divisible by 6.


Let P(k) = k(k2 + 5) is divisible by 6.; k(k2 + 5) = 6x.


P(k+1) = (k+1)((k+1)2 + 5) = (k+1)(k2+2k+6)


= k3 + 3k2 + 8k + 6


= 6x+3k2+3k+6


= 6x+3k(k+1)+6[n(n+1) is always even and divisible by 2]


= 6x + 3×2y + 6 ; is divisible by 6.


P(k+1) is true when P(k) is true.


By Mathematical Induction P(n) = n(n2 + 5) is divisible by 6, for each natural number n.


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