Difference between revisions of "TADM2E 1.16"
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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 leftside part equals <math>a</math>, and the rightside is clearly divisible by 3, thus the <math>n^3 + 2n</math> is proven to be divisible by 3.