mathematical induction

Let P(n): n^3+5n is divisible by 6 for n is positive integer

For n =1
(1)^3+5(1) =1+ 5 = 6 is divisible by 6

Assume P(k) is true
k^3+5k = 6m for some m is an integer

(k+1)^3+5(k+1)
=k^3+3k^2+3k+1+5k+5
=(k^3+5k)+3k(k+1)+6 (k(k+1) is an even no.)
=6m +6n +6 (n=[k(k+1)/2 is an integer)
=6(m+n+1) is divisible by 6
P(k+1) is also true.

By MI, P(n) is true for all n is a positive integer

Prove by mathematical induction that n³+5n is divisible by 6 for any positive integers n.
if n=1
1³+5(1)=6
which is divisible by 6
Inductive step:
Assume that n³+5n is divisible by 6 for any positive integers n
(n³+5n)/6=m
(1/6)(n)(n²+5)=m, where m is positive integers
we show
((n+1)³+5(n+1))/6=p, where p is positive intergers
(1/6)(n+1)((n+1)²+5)=(1/6)(k)(k²+5)=p, where k and p are positive intergers
Basis step and the Inducive step have been verified.

let p(n) be the proposition
n^3+5n for all integers
when n=1
1^3+5(1)=6(1)

thus p(n) is true for n=1

Assume p(n) is true for n=k,where k is a positive integer
i.e.: k^3+5k=6m m: integer
For n=k+1
(k+1)^3+5(k+1)

之後唔識做 ^^