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

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.

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

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:

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