# TADM2E 1.16

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Revision as of 18:22, 11 September 2014 by Algowikiadmin (Talk | contribs)

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.