# Difference between revisions of "TADM2E 1.16"

First we must assume that $\frac {n^3 + 2n}{3} = a$, where $a \in N$.
The case when $n = 0$ is pretty obvious, so let's go straight to $n = n + 1$:
We get: $\frac {(n + 1)^3 + 2(n + 1)}{3} = \frac { n^3 + 3n^2 + 3n + 1 + 2n + 2 }{3}$.
Now let's rearrange the parts like this: $\frac {n^3 + 2n}{3} + \frac {3n^2 + 3n + 3}{3}$, and we can see, that the left-side part equals $a$, and the right-side is clearly divisible by 3, thus the $n^3 + 2n$ is proven to be divisible by 3.