mathematical induction

Let f(n) = n^3+5n

(1)
For n=1, f(1) = 6 divisible by 6.
f(n) is true for n=1

(2)
For n=2, f(2) = 18 divisible by 6.
f(n) is true for n=2

(3)
For n=k is true, where k is positive integer,
Let f(k) = k^3 + 5k = 6p, where p is positive integer,
f(k+2)
= (k+2)^3 + 5(k+2)
= k^3 + 6k^2 + 17k + 18
= 6p + 6k^2 + 12k + 18
= 6 (p + k^2 + 2k + 3) divisible by 6

By mathematical induction, f(n) is divisible by 6.

n = 1,
n^3+5n
= 1+ 5
= 6 which is divisible by 6

Assume n^3 + 5n is divisible by 6 when n = k
ie. k^3 + 5k = 6r where r is an integer

when n = k + 1
n^3+5n
= (k+1)^3 + 5 (k+1)
= k^3 + 3k^2 + 3k + 1 + 5k + 5
= k^3 + 5k + 3k^2+3k+6
= 6r + 3(k^2+k+2)
= 6r + 6+3k(k+1)

Since k(k+1) are consecutive integers
one of them must be odd number and another one must be even number
the product of odd number and even number must be even number
Therefore k(k+1) = 2s where s is an integer

6r+6+3k(k+1)
= 6r+6+6s
= 6(r+s+1) which is also divisible by 6