I need to prove: n^{3 }+ 5n is divisible by 6.

n=1

1^{3}+5(1)=6 which is divisible by 6 = true.

n=2

2^{3}+5(2)=18 which is divisible by 6 = true.

What else do I need to prove?

- Apr 6th 2013, 06:32 AMrtrumpowUse mathematical induction to prove the truth assertions for all n≥1
I need to prove: n

^{3 }+ 5n is divisible by 6.

n=1

1^{3}+5(1)=6 which is divisible by 6 = true.

n=2

2^{3}+5(2)=18 which is divisible by 6 = true.

What else do I need to prove? - Apr 6th 2013, 06:53 AMPlatoRe: Use mathematical induction to prove the truth assertions for all n≥1
Suppose that $\displaystyle K^3+5K$ is divisble by six, where $\displaystyle K\ge 3$ is a positive integer.

Using that, prove that__fact__

$\displaystyle \begin{align*} (K+1)^3+5(K+1)&=(K^3+3K^2+3K+1)+(5K+5)\\&=(K^3+5K) +3K(3K+1)+6\end{align*}$

is divisble by six.

HINT: The sum of three multiples of six is divisble by six. - Apr 6th 2013, 09:24 AMHallsofIvyRe: Use mathematical induction to prove the truth assertions for all n≥1
I'm puzzled as to why you were given a problem requiring proof by induction if, as you seem to be saying, you have no idea what "proof by induction" means.

- Apr 8th 2013, 07:43 PMmathguy25Re: Use mathematical induction to prove the truth assertions for all n≥1
Remember that steps to induction goes like this

Let P be a statement. Show that

P(1) is true

If P(k) is true, then P(k+1) is true

You already showed that P(1) is true. Now just need to show P(k+1) is true if P(k) is true. Let P(k) be true. Then k^3 + 5k is divisible by 6. Need to show that (k+1)^3 + 5(k+1) is divisible by 6. Expand.

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

Clearly 6 is divisible by 6. By induction assumption k^3 + 5k is divisible by 6. I claim that 3k(k+1) is divisible by 6. That is because k(k+1) is even for all k. If k is even, then clearly k(k+1) is even so 3k(k+1) is divisible by 6. If k is odd, then k + 1 is even so k(k+1) is even and again 3k(k+1) is divisible by 6. The sum of terms each of which are divisible by 6 is also divisible by 6. - Apr 8th 2013, 09:03 PMGusbobRe: Use mathematical induction to prove the truth assertions for all n≥1
I know the OP probably needed to use induction. For completeness, I'll give a proof without induction. I hope this gives more intuition as to why the statement is true.

**Lemma**

$\displaystyle n^3+5n$ is divisible by 6 if and only if $\displaystyle n^3-n$ is divisible by 6.

Proof: Suppose $\displaystyle n^3+5n=6k$, then $\displaystyle n^3-n=6(k-n)$. $\displaystyle k>n$ because $\displaystyle k= \frac{n(n^2+5)}{6}\geq n$ since $\displaystyle n^2+5 \geq 6$ for $\displaystyle n\in \mathbb{N}$. Conversely, $\displaystyle n^3-n=6k \Rightarrow n^3+5n=6(k+n)$.

**Proposition**

$\displaystyle n^3+5n$ is divisible by 6.

Proof: From the lemma, the proposition holds iff $\displaystyle n^3-n$ is divisible by 6. But $\displaystyle n^3-n=n(n^2-1)=(n-1)(n)(n+1)$. These are three consecutive integers, and so one of them must be divisible by 3.