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Math rule I don't understand.

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

peterh
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Irresponsible Newb
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4 Answers4

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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.

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

David Richerby
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NoName
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    Hooray. A simple equality about arithmetic proved using simple arithmetic, instead of by invoking geometric progressions or base-7. – David Richerby Oct 22 '14 at 20:31
  • “$(7^{n+1}-1)/6 + 7^{n+1} = (7^{n+2}-1)/6$” is not an equation; you misunderstand the latter term. – Incnis Mrsi Oct 23 '14 at 08:02
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    @IncnisMrsi No, you misunderstand the term "equation". An equation is any two things related by $=$. – David Richerby Oct 23 '14 at 08:06
  • @David Richerby keywords: equality, binary relation, (arithmetic) identity, “algebra vs arithmetic”. – Incnis Mrsi Oct 23 '14 at 08:16
  • @IncnisMrsi I have no idea why you're listing "keywords" at me. If you believe I have given the wrong definition of "equation", please post the correct definition so that people can learn, instead of just claiming that I'm wrong and insulting me, which has no place here. In this case, if you look up the definition of "equation", you'll see that I'm correct. – David Richerby Oct 23 '14 at 08:47
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    @DavidRicherby It's an equality. It would be an equation if $n$ was an unknown we wanted to solve for. Here, it's not. – Angew is no longer proud of SO Oct 23 '14 at 09:17
  • @Angew The Wikipedia page for "equation" which you link to says that an equation "may contain one or several... unknowns" (emphasis mine). This is an equation that happens not to contain unknowns. – David Richerby Oct 23 '14 at 09:31
  • @DavidRicherby And quoting the "equality" page: "An equation is the problem of finding values of some variables, called unknowns, for which the specified equality is true." You can probably define an "unknownless equation," but I've always heard the equality vs. equation distinction as relation vs. problem. – Angew is no longer proud of SO Oct 23 '14 at 09:35
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    I was one of those who wrote the definition of equation in Wikipedia ☺☺ – Incnis Mrsi Oct 23 '14 at 09:36
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    @Angew So the two pages you quote show that there is no consensus that the definition of "equation" that you are using is the only correct definition. So you're wrong to say that I'm wrong, since I'm giving a definition that is widely accepted as correct. – David Richerby Oct 23 '14 at 09:38
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    Overall this argument doesn't matter at all. How pointless. –  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 $$ \frac{7^k-1}{6} = \underbrace{11\ldots 11}_{k\text{ ones}}{}_7 $$

hmakholm left over Monica
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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:

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

MJD
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    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, $(7^{n+1}-1)/6 + 7^{n+1} = (7^{n+1}+6\times 7^{n+1}-1)/6 = (7\times 7^{n+1}-1)/6 = (7^{n+2}-1)/6$. – David Richerby Oct 22 '14 at 20:37
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    Unnecessarily complex, not sure why this is the selected answer – Math_Illiterate Oct 23 '14 at 04:23
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    @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. – Michael Shaw Oct 23 '14 at 05:43
  • @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? – David Richerby Oct 23 '14 at 08:08
  • @DavidRicherby: It's not just "can I prove it?" but also "can I remember what it is?" and "can I understand it?". If you need to use this rule (or something like it) to prove something else, knowing that a little arithmetic would prove the rule that you can't quite remember won't help you remember it. Understanding it would let you reconstruct it from scratch, and you wouldn't need to remember it. – Michael Shaw Oct 23 '14 at 10:19
  • @MichaelShaw The only time you need to use the "rule" is when you have an expression like $(a^n-b)/(a-1) + a^n$ in front of you. You don't need to remember any rules: just do the arithmetic on the expression you see. – David Richerby Oct 23 '14 at 10:26
  • @DavidRicherby: Understanding why the rule is true is a lot broader than that one expression. Reading the question, it seems to me like he's looking for an understanding of why the book mentioned it at all, what it's for, what's behind it, etc. Maybe I'm reading too much into it, but he did select this answer, so if he wasn't looking specifically for it, he at least appreciated it. – Michael Shaw Oct 23 '14 at 11:20
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    @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. – MJD Oct 23 '14 at 13:01
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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} $

Adilson de Almeida Jr
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    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. – John Feminella Oct 23 '14 at 00:19
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    @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! – Adilson de Almeida Jr Oct 23 '14 at 02:06
  • no magic, no drama. very simple indeed – prusswan Oct 23 '14 at 05:42