Difference between revisions of "TADM2E 1.16"

Revision as of 18:13, 11 September 2014

First we must assume that <math>\frac {n^3 + 2n}{3} = a</math>, where <math>a \in N</math>.

The case when <math>n = 0</math> is pretty obvious, so let's go straight to <math>n = n + 1</math>:

We get: <math>\frac {(n + 1)^3 + 2(n + 1)}{3} = \frac { n^3 + 3n^2 + 3n + 1 + 2n + 2 }{3}</math>.

Now let's rearrange the parts like this: <math>\frac {n^3 + 2n}{3} + \frac {3n^2 + 3n + 3}{3}</math>, and we can see, that the left-side part equals <math>a</math>, and the right-side is clearly divisible by 3, thus the <math>n^3 + 2n</math> is proven to be divisible by 3.