# Confusing algebra rule: why $\frac{7^{n+1}-1}{6} + 7^{n+1} = \frac{7^{n+2}-1}{6}$?

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

• The secret rule applied is $1+6=7$ (after bringing things to a common denominator). Oct 23, 2014 at 9:32

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}

• Hooray. A simple equality about arithmetic proved using simple arithmetic, instead of by invoking geometric progressions or base-7. Oct 22, 2014 at 20:31
• @IncnisMrsi No, you misunderstand the term "equation". An equation is any two things related by $=$. Oct 23, 2014 at 8:06
• @DavidRicherby It's an equality. It would be an equation if $n$ was an unknown we wanted to solve for. Here, it's not. Oct 23, 2014 at 9:17
• I was one of those who wrote the definition of equation in Wikipedia ☺☺ Oct 23, 2014 at 9:36
• Overall this argument doesn't matter at all. How pointless.
– user142198
Oct 23, 2014 at 10:04

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 $$\frac{7^k-1}{6} = \underbrace{11\ldots 11}_{k\text{ ones}}{}_7$$

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:

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

• 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, $(7^{n+1}-1)/6 + 7^{n+1} = (7^{n+1}+6\times 7^{n+1}-1)/6 = (7\times 7^{n+1}-1)/6 = (7^{n+2}-1)/6$. Oct 22, 2014 at 20:37
• Unnecessarily complex, not sure why this is the selected answer Oct 23, 2014 at 4:23
• @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. Oct 23, 2014 at 5:43
• @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? Oct 23, 2014 at 8:08
• @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, 2014 at 13:01

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:

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

• 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. Oct 23, 2014 at 0:19
• @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! Oct 23, 2014 at 2:06
• no magic, no drama. very simple indeed Oct 23, 2014 at 5:42