# Difference between revisions of "TADM2E 1.16"

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

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

− | We get: | + | 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: | 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. |

## Latest revision as of 18:22, 11 September 2014

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.