Algowikiadmin: Recovering wiki

2014-09-11T18:22:55Z

Recovering wiki

Algowikiadmin: Recovering wiki

2014-09-11T18:13:36Z

Recovering wiki

New page

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.

@@ Line 1: / Line 1: @@
-First we must assume that &lt;math&gt;\frac {n^3 + 2n}{3} = a&lt;/math&gt;, where &lt;math&gt;a \in N&lt;/math&gt;.
+First we must assume that <math>\frac {n^3 + 2n}{3} = a</math>, where <math>a \in N</math>.
-The case when &lt;math&gt;n = 0&lt;/math&gt; is pretty obvious, so let's go straight to &lt;math&gt;n = n + 1&lt;/math&gt;:
+The case when <math>n = 0</math> is pretty obvious, so let's go straight to <math>n = n + 1</math>:
-We get: &lt;math&gt;\frac {(n + 1)^3 + 2(n + 1)}{3} = \frac { n^3 + 3n^2 + 3n + 1 + 2n + 2 }{3}&lt;/math&gt;.
+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:
-&lt;math&gt;\frac {n^3 + 2n}{3} + \frac {3n^2 + 3n + 3}{3}&lt;/math&gt;, and we can see, that the left-side part equals &lt;math&gt;a&lt;/math&gt;, and the right-side is clearly divisible by 3, thus the &lt;math&gt;n^3 + 2n&lt;/math&gt; is proven to be divisible by 3.
+<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.

TADM2E 1.16 - Revision history

Algowikiadmin: Recovering wiki

Algowikiadmin: Recovering wiki