# Thread: prove that n^3 + 5n is divisble by 6 using induction

1. ## prove that n^3 + 5n is divisble by 6 using induction

So I started doing this proof then realized I hasn't really used induction. I said that n^3 + 5n is equal to n^3 - n + 6n = n(n^2 - 1) + 6n = n(n-1)(n+1) + 6n = (n-1)(n)(n+1) + 6n

Now I know that 6n is divisible by 6 and the product of 3 consecutive integers is divisible by 6 so that the entire term is divisble by 6 but I don't know how to write a proper proof using that explanation. I know you can do this using induction although im not sure how. Thanks for your help

2. Well, what's your base case going to be?

3. Originally Posted by mremwo
So I started doing this proof then realized I hasn't really used induction. I said that n^3 + 5n is equal to n^3 - n + 6n = n(n^2 - 1) + 6n = n(n-1)(n+1) + 6n = (n-1)(n)(n+1) + 6n

Now I know that 6n is divisible by 6 and the product of 3 consecutive integers is divisible by 6 so that the entire term is divisble by 6 but I don't know how to write a proper proof using that explanation. I know you can do this using induction although im not sure how. Thanks for your help
Exactly one of every three consecutive integers is divisible by 3. Same goes for 2. (Except that's every second consecutive integer.)

4. Originally Posted by mremwo
So I started doing this proof then realized I hasn't really used induction. I said that n^3 + 5n is equal to n^3 - n + 6n = n(n^2 - 1) + 6n = n(n-1)(n+1) + 6n = (n-1)(n)(n+1) + 6n

Now I know that 6n is divisible by 6 and the product of 3 consecutive integers is divisible by 6 so that the entire term is divisble by 6 but I don't know how to write a proper proof using that explanation. I know you can do this using induction although im not sure how. Thanks for your help
Since this works $\displaystyle \forall n\geq 1$

Show P(1) is true.

Then let $\displaystyle n\leq k$ where k is an arbitrary integer.

Assume P(k) is true and then prove P(k+1) is true.

5. Mremwo, your proof looks good to me. It's just not a proof by induction.

Your proof is actually better than an inductive proof because it proves the statement for all integers n. A proof by induction needs a starting point.

6. Originally Posted by mremwo
So I started doing this proof then realized I hasn't really used induction. I said that n^3 + 5n is equal to n^3 - n + 6n = n(n^2 - 1) + 6n = n(n-1)(n+1) + 6n = (n-1)(n)(n+1) + 6n

Now I know that 6n is divisible by 6 and the product of 3 consecutive integers is divisible by 6 so that the entire term is divisble by 6 but I don't know how to write a proper proof using that explanation. I know you can do this using induction although im not sure how. Thanks for your help
P(k)

The proposition for n=k is

$\displaystyle k^3+5k$ is divisible by 6

P(k+1)

The proposition for the "next n" is

$\displaystyle (k+1)^3+5(k+1)$ is divisible by 6.

You need to prove that the Induction Domino-Effect exists
by showing that P(k+1) will be true "if" P(k) is true.

Hence, either write P(k+1) in terms of P(k),
when we will see if P(k+1) will be true if P(k) is..

Or work towards P(k+1) from P(k).

Proof

$\displaystyle (k+1)^3+5(k+1)=k^3+3k^2+3k+1+5k+5$

$\displaystyle =\left(k^3+5k\right)+3k^2+3k+6=\left(k^3+5k\right) +3k(k+1)+6$

Since k and k+1 are a pair of consecutive even and odd terms, $\displaystyle 3k(k+1)$ is a multiple of 6 for $\displaystyle k \ge\ 1$

Then we can see if P(k+1) will be true "if" P(k) is.

Prove for a base case to topple the dominoes.

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# n^3 5n divisible by 6 ans

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