F.4 A Maths M.I.

Let S(n) be the statement n^3+5n is divisible by 6 for all natural numbers n.

When n=1
LHS = 1^3 + 5*1 = 6 which is divisible by 6.

S(1) is true.

Assume S(k) is true, i.e. k^3+5k is divisible by 6 or k^3+5k = 6j for some natural numbers j

When n = k+1
(k+1)^3+5(k+1)
=k^3+3k^2+3k+1+5k+5
=k^3+5k+3k^2+3k+6
=6j+3k(k+1)+6

when k is an odd number, (k+1) is an even number
vice versa, when k is an even number, (k+1) is an odd number
therefore, k(k+1) can be written as 2p for some natural numbers p

therefore LHS = 6j+3*2p+6
=6j+6p+6
=6(j+p+1)
which is divisible by 6.

S(k+1) is true

By MI, S(n) is true for all natural numbers n.