n(n² + 5) is divisible by 6, for each natural number n.

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.