Can you please help me with
Prove that n^3-n is divisible by 3
$\displaystyle n^3 - n = 3a$ for some $\displaystyle a$.
First we see if it holds for $\displaystyle n=0$
$\displaystyle 0^3 - 0 = 0$, therefore true for $\displaystyle n=0$
Now we use what's called the induction step, we assume that it's true for $\displaystyle n=k$ where k is some natural nubmer.
This gives us $\displaystyle k^3 - k = 3a$
We then prove this for $\displaystyle n=k+1$, giving us
$\displaystyle (k+1)^3 - k - 1$.
Expanding the brackets gives $\displaystyle k^3 + 3k^2 + 3k + 1 - k - 1$
$\displaystyle k^3 + 3k^2 + 2k$ which can be written as $\displaystyle k^3 + 3k^2 + 3k - k$
$\displaystyle (k^3-k) + 3(k^2 + k) = 3a + 3(k^2 + k)$.
Therefore it is true for $\displaystyle k+1$ if true for $\displaystyle k$.
You know it is true for 0, therefore true for 1. True for one, therefore true for 2....
Hope this helps