Originally Posted by

**adkinsjr** I want to prove the following theorem:

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If $\displaystyle n$ is a positive integer, then $\displaystyle 7^n-1$ is divisible by 6.

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The only idea I have so far is the following equation:

$\displaystyle \frac{1}{6}(7^n-1)=a_n$

Where $\displaystyle a_n$ is an infinite sequence:

$\displaystyle a_n={1,8,57,400,2801,...}$

The sequence doesn't have a common ratio, however I have observed the following:

$\displaystyle 8:1=8$

$\displaystyle 57:8=7.125$

$\displaystyle 400:57\approx 7.02$

$\displaystyle 2801:400=7.0025$

So it seems that $\displaystyle \frac{1}{6}\lim_{n->\infty}\frac{7^n-1}{7^{n-1}-1}=7$

I'm not really sure if the limit is useful or not.