prove using mathematical induction, `n(n+1)(n+5)` is divisible by `6` for all natural numbers

Explanation

Let `P(n)=n(n+1)(n+5)`
Step I For `n=1`,
`P(1)=1.(1+1)(1+5)=1.2.6=12`, which is divisible by 6.
Therefore , there result is true for `n=1`.
Step II Assume that the result is true for `n=k` Then , `{:(" "P(k)=k(k+1)(k+5),"is divisible by",6.),(rArrP(k)=6r", "r "is an interger",,):}}`
Step III For `n=k+1`.
`P(k+1)=(k+1)(k+1+1)(k+1+5)=(k+1)(k+2)(k+6)`
Now, `P(k+1)-P(k)=(k+1)(k+2)(k+6) -k(k+1)(k+5)`
`=(k+1){k^2+8k+12-k^2-5k}`
`-=(k+1)(3k+12)`
`=3(k+1)(k+4)`
`rArr P(k+1)=P(k)+3(k+1)(k+4)`
which is divisible by 6 as P(k) is divisible by 6 [by assumption step]
and clearly , `3(k+1)(k+4)` is divisible by `6 forall , k in N`.
Hence , the result is true for `n=k+1`.
Therefore , by the principle of mathematical induction , the result is true for all `n in N`.