1. ## Another proof question

I want to prove the following theorem:
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If $n$ is a positive integer, then $7^n-1$ is divisible by 6.
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The only idea I have so far is the following equation:

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

Where $a_n$ is an infinite sequence:

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

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

$8:1=8$
$57:8=7.125$
$400:57\approx 7.02$
$2801:400=7.0025$

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

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

Prove the following:

If $n$ is a positive integer, then $7^n-1$ is divisible by 6.
Consider the expansion of $(6+1)^n$

. . $7^n \;\;=\;\;(6\;+\;1)^n \;=\; \;6^n \;+\; {n\choose1}6^{n-1} \;+\; {n\choose2}6^{n-2} \;+ \hdots +\; {n\choose n-2}6^2 \;+\; {n\choose n-1}6 \;+\; 1$

Then: . $7^n - 1 \;\;=\;\;6^n + {n\choose1}6^{n-1} + {n\choose2}6^{n-2} + \hdots + {n\choose n-2}6^2 + {n\choose n-1}6$

. . . . . . . . . . $=\;\;6\cdot\underbrace{\bigg[6^{n-1} + {n\choose1}6^{n-2} + {n\choose2}6^{n-3} + \hdots + {n\choose n-2}6 + {n\choose n-1}\bigg]}_{\text{This is an integer}}$

Hence, $7n - 1$ is a multiple of 6.

Therefore, $7^n-1$ is divisible by 6.

3. Originally Posted by Soroban

Consider the expansion of $(6+1)^n$

. . $7^n \;\;=\;\;(6\;+\;1)^n \;=\; \;6^n \;+\; {n\choose1}6^{n-1} \;+\; {n\choose2}6^{n-2} \;+ \hdots +\; {n\choose n-2}6^2 \;+\; {n\choose n-1}6 \;+\; 1$

Then: . $7^n - 1 \;\;=\;\;6^n + {n\choose1}6^{n-1} + {n\choose2}6^{n-2} + \hdots + {n\choose n-2}6^2 + {n\choose n-1}6$

. . . . . . . . . . $=\;\;6\cdot\underbrace{\bigg[6^{n-1} + {n\choose1}6^{n-2} + {n\choose2}6^{n-3} + \hdots + {n\choose n-2}6 + {n\choose n-1}\bigg]}_{\text{This is an integer}}$

Hence, $7n - 1$ is a multiple of 6.

Therefore, $7^n-1$ is divisible by 6.
Ok, I understand this. But how did you develop the idea of expanding the binomial? When ever I go to prove theorems like this, I'm not sure where to begin. You're proof just isn't something I would have noticed immediately.

What particular branch of mathematics should I study in order to obtain this skill?

4. ## Another Proof Question

I am not the one that helped you with your proof, but I am a Mathematics major. Depending on how far you plan to go with developing proofs, you may want to take Discrete Math. If you plan to pursue a career in Mathematics you may want to take something like Foundations of Mathematics. The college you attend may call it something else, but it will be a class strictly on writing proofs. Talk to your adviser and find out what the best classes to take relating to your major.

Hope you find what you are looking for!

I want to prove the following theorem:
----------------------------------------
If $n$ is a positive integer, then $7^n-1$ is divisible by 6.
----------------------------------------
The only idea I have so far is the following equation:

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

Where $a_n$ is an infinite sequence:

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

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

$8:1=8$
$57:8=7.125$
$400:57\approx 7.02$
$2801:400=7.0025$

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

I'm not really sure if the limit is useful or not.
It's useful to know that $a^n-b^n=(a-b)\cdot(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots+ab^{n-2}+b^{n-1})$, since if you multiply out it's easy to see that all terms cancel except $a\cdot a^{n-1}$ and $-b\cdot b^{n-1}$.
This can be applied to your problem directly like this $7^n-1=7^n-1^n=(7-1)\cdot(7^{n-1}+7^{n-2}\cdot 1+\cdots+7\cdot 1^{n-2}+1^{n-1})$