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
 A: 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.
$$
\begin{align}
\frac{7^{n+1} - 1}{6} + 7^{n+1}
&= \frac{7^{n+1} - 1}{6} + \frac{6\cdot 7^{n+1}}{6} \\
&= \frac{7^{n+1} - 1 + 6 \cdot 7^{n+1}}{6} \\
&= \frac{7 \cdot 7^{n+1} - 1}{6} \\
&= \frac{7^{n+2} -1}{6}.
\end{align}
$$
A: 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} $
A: 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
$$ \frac{7^k-1}{6} = \underbrace{11\ldots 11}_{k\text{ ones}}{}_7 $$
A: 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:
$$\color{maroon}{\frac{7^{n+1}-1}{6}}  + 7^{n+1} = \color{darkblue}{\frac{7^{n+2}-1}{6}}$$
