# Using mathematical induction, prove that n^3+2n is divisible by 3 for all integers n

To prove this we must **use a neat mathematical technique called induction.**

Induction works in the following way: If you show that the result being true for any integer implies it is true for the next, then you need only show that it is true for n=1 for it to be true for n=2 and then n=3 and so on.

__Step 1: Show true for n=1__

For n=1, n^3+2n=(1)^3+2(1)

n^3+2n=3

**3 is definitely divisible by 3 so the statement is true for n=1.**

**Step 2: Assume true for n=k**

We assume that for any integer k, n^3+2n is divisible by 3. We can write this mathematically as:

**k^3+2k=3m, where m is an integer**

__Step 3: Show true for k+1__

For n=k+1,

n^3+2n=(k+1)^3+2(k+1)

=(k^3+3k^2+3k+1)+2k+2

=(k^3+2k)+3(k^2+k+1)

**Subbing in from part 2 for (k^3+2k), **we get:

n^3+2n=3m+3(k^2+k+1)

=3(m+k^2+k+1)

**which is divisible by 3.**

**This means that the statement being true for n=k implies the statement is true for n=k+1, and as we have shown it to be true for n=1 the proof of the statement follows by induction.**