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

REPLY [4 votes]: 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} $