How to demonstrate this with mathematical induction?`P(n):n^3+5n` divide with 6 is true `forall ninNN`.

To prove this identity using mathematical induction we have to follow these steps:

1. Check the identity for `n=1`: `P(1)=1^3+5*1=6`. The result is a multiple of `6`, so the thesis is true.
2. Assume that it is true for `n=k`, so: `EE{a in ZZ} k^3+5k=6a`
3. Next step is to prove (using the assumption) that the thesis is true for `n=k+1`

Proof

`n=k+1`

`(k+1)^3+5(k+1)=k^3+3k^2+3k+1+5k+5=`
`=k^3+5k+3k^2+3k+6`

Now we can simplify the first 2 term using the assumption from point 2.

`k^3+5k+3k^2+3k+6=6a+3k^2+3k+6=`

`6a+3(k^2+k+2)=6a+3(k(k+1)+2)`

The first component of the sum is divisible by `6` (from the assumption). We have to show that the second expression is also divisible by `6`.

It is clearly divisible by `3`. As a sum of an even number and a product of 2 consecutive natural numbers the expression in brackets is an even number. Therfore the whole expression is divisible by `6`.

This concludes the proof.

1) test for P(1)

`P(1):1^3+5xx1=1+5=6`

`:.P(1)=0("mod 6")`

`:." true for " n=1`

2) Assume true `n=k`

`P(k)=k^3+5k`

3) Prove for `k+1`

`(k+1)^3+5(K+1)=k^3+3k^2+3k+1+5k+5`

`P(k+1)=k^3+3k^2+8k+6`

`=>P(k+1)=(k^3+5k)+3(k^2+k++6)`

`(k^3+5k)=6p " by assumption"`

`=>P(k+1)=6p+3(k^2+k+6)`

`3(k^3+k++6)=0 ( mod 3)`

`(k^3+k+6) " for k "odd`

`k=2q-1=>(k^2+k+6)=(2q-1)^2+(2q-1)+6`

`=4q^2+2q^2+6=2(2q^2+2q^2+3)=0(mod2)`

so`" " 3(k^2+k+6)=6m`

`:.P(k+1) =6(p+m)=0(mod6) " for k "odd"`

`k " even"=k=2t`

`=>(k^2+k+6)=(2t)^2+2t+6=4t^2+2t+6`

`=2(2t^2+t+3)=2s`

`:.P(k)=6(p+s)=0(mod6) " for k even"`

`P(k)=>P(k+1) AAkinNN`

but true for `P(1)`

`:." by induction " P(1)=>P(2)=>P(3)....`

`" so true "AAninNN`

To compare, a non inductive proof

We know that

`n(n+1)(n+2)` is divisible by `6` because one of `n, n+1, n+2` is divisible by `3` and also at least one of them is even.

so `n(n+1)(n+2) equiv 0 mod 6`

but

`n^3+5n = n(n+1)(n+2) -3n(n-1)`

and here `3n(n-1) equiv 0 mod 6`

so

`n^3+5n equiv 0 mod 6`