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Math rule I don't understand.

Hey guys, my discrete math midterm is tomorrow and I'm studying proof styles. I came across a rule (algebra maybe?) I don't quite understand and I was hoping someone could explain it step by step for me.

$$\frac{7^{n+1}-1}{6} + 7^{n+1} = \frac{7^{n+2}-1}{6}$$

I guess I can memorize it, but could someone show me how it works step by step?

Thanks

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    $\begingroup$ The secret rule applied is $1+6=7$ (after bringing things to a common denominator). $\endgroup$ – Marc van Leeuwen Oct 23 '14 at 9:32
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Do you know the rule for the sum of a finite geometric series?

$$1 + a + a^2 + \cdots + a^n = \frac{a^{n+1}-1}{a-1}$$

Now take $a=7$:

$$\begin{align} 1 + 7 + 7^2 + \cdots + 7^n\hphantom{+7^{n+1}} &= \color{maroon}{\frac{7^{n+1}-1}{6}} \\ 1 + 7 + 7^2 + \cdots + 7^n+7^{n+1} &= \color{darkblue}{\frac{7^{n+2}-1}{6}} \\ \end{align} $$

The second line is the same as the first line, but with $7^{n+1}$ added:

$$\color{maroon}{\frac{7^{n+1}-1}{6}} + 7^{n+1} = \color{darkblue}{\frac{7^{n+2}-1}{6}}$$

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    $\begingroup$ This is very clever and quite beautiful but it's also the sort of thing that convinces people that they can't do maths because you need to be some kind of genius. Come on, $(7^{n+1}-1)/6 + 7^{n+1} = (7^{n+1}+6\times 7^{n+1}-1)/6 = (7\times 7^{n+1}-1)/6 = (7^{n+2}-1)/6$. $\endgroup$ – David Richerby Oct 22 '14 at 20:37
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    $\begingroup$ Unnecessarily complex, not sure why this is the selected answer $\endgroup$ – Math_Illiterate Oct 23 '14 at 4:23
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    $\begingroup$ @Math_Illiterate: It shows why they're equal, in a more informative way than just calculating along. If you look at the question, he was considering memorizing it and wanted to know if there was a meaning he could understand so he could avoid memorizing a bunch of symbols. This answer (and the one about base-7) explain the meaning behind the equation. If you put both answers together, the rule for a geometric series makes sense in a way that makes it unnecessary to memorize as well. Both answers are conceptually very simple. $\endgroup$ – Michael Shaw Oct 23 '14 at 5:43
  • $\begingroup$ @MichaelShaw It's true because it follows immediately from the laws of arithmetic. Why bother memorizing "I can prove this equality using geometric series or base-7 representations" when you can prove it in a couple of lines of ordinary arithmetic? $\endgroup$ – David Richerby Oct 23 '14 at 8:08
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    $\begingroup$ @david I don't think it's excessively clever or complex; if I had I wouldn't have posted it. When I saw the form of the expressions in the question, I thought immediately of the sum of a finite geometric series, which is part of the high-school curriculum in the United States. I thought it was quite possible that the OP was studying this topic in school right now and that the geometric series solution was the whole point of the question. I think it's often a good idea to point out “this complicated-seeming expression is actually something you have seen before”, so that's what I did. $\endgroup$ – MJD Oct 23 '14 at 13:01
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This isn't the sort of rule you need to memorize, but you do need know the operations to get from one side of the equation to the other.

$$ \begin{align} \frac{7^{n+1} - 1}{6} + 7^{n+1} &= \frac{7^{n+1} - 1}{6} + \frac{6\cdot 7^{n+1}}{6} \\ &= \frac{7^{n+1} - 1 + 6 \cdot 7^{n+1}}{6} \\ &= \frac{7 \cdot 7^{n+1} - 1}{6} \\ &= \frac{7^{n+2} -1}{6}. \end{align} $$

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    $\begingroup$ Hooray. A simple equality about arithmetic proved using simple arithmetic, instead of by invoking geometric progressions or base-7. $\endgroup$ – David Richerby Oct 22 '14 at 20:31
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    $\begingroup$ @IncnisMrsi No, you misunderstand the term "equation". An equation is any two things related by $=$. $\endgroup$ – David Richerby Oct 23 '14 at 8:06
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    $\begingroup$ @DavidRicherby It's an equality. It would be an equation if $n$ was an unknown we wanted to solve for. Here, it's not. $\endgroup$ – Angew Oct 23 '14 at 9:17
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    $\begingroup$ I was one of those who wrote the definition of equation in Wikipedia ☺☺ $\endgroup$ – Incnis Mrsi Oct 23 '14 at 9:36
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    $\begingroup$ Overall this argument doesn't matter at all. How pointless. $\endgroup$ – user142198 Oct 23 '14 at 10:04
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Think base 7. Then your rule says $$ \underbrace{11\ldots11}_{n+1\text{ ones}}{}_7 + 1\underbrace{00\ldots 00}_{n+1\text{ zeroes}}{}_7 = \underbrace{11\ldots11}_{n+2\text{ ones}}{}_7 $$ because $$ \frac{7^k-1}{6} = \underbrace{11\ldots 11}_{k\text{ ones}}{}_7 $$

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This is really very simple.

$ (7^{n+1}−1)/6+7^{n+1} $

putting it on the same denominator:

$ ... = \dfrac{7^{n+1}}{6} - \dfrac{1}{6} + \dfrac{6 \times 7^{n+1}}{6} $

grouping the first and the last term:

$ ... = \dfrac{ 7 \times 7^{n+1}}{6} - \dfrac{1}{6} $

doing the last multiplication:

$ ... = \dfrac{ 7^{n+2}}{6} - \dfrac{1}{6} = \dfrac{ 7^{n+2} - 1}{6} $

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    $\begingroup$ I don't think telling people it's "very simple" is helpful; it increases the feeling of discouragement if they don't get it, like saying something is "obvious" or "trivial". Show, don't tell. $\endgroup$ – John Feminella Oct 23 '14 at 0:19
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    $\begingroup$ @JohnFeminella When I said it's very simple, i would mean it's a problem which can be solved without any previous knowledge of series, he could use only the high school math. My apologies if I sounded pedantic. Anyway, thanks for the advice! $\endgroup$ – Adilson de Almeida Jr Oct 23 '14 at 2:06
  • $\begingroup$ no magic, no drama. very simple indeed $\endgroup$ – prusswan Oct 23 '14 at 5:42

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