Answer :

Given; P(n) = n(n^{2} + 5) is divisible by 6.

P(0) = 0(0^{2} + 5) = 0 ; is divisible by 6.

P(1) = 1(1^{2} + 5) = 6 ; is divisible by 6.

P(2) = 2(2^{2} + 5) = 18 ; is divisible by 6.

P(3) = 3(3^{2} + 5) = 42 ; is divisible by 6.

Let P(k) = k(k^{2} + 5) is divisible by 6.; ⇒ k(k^{2} + 5) = 6x.

⇒ P(k+1) = (k+1)((k+1)^{2} + 5) = (k+1)(k^{2}+2k+6)

= k^{3} + 3k^{2} + 8k + 6

= 6x+3k^{2}+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(n^{2} + 5) is divisible by 6, for each natural number n.

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