Not a question, just a useful formula

This can be generalized:

$a^0 + a^1 + \dots + a^n = \frac{1 - a^{n+1}}{1 - a}$

(I derived this from the formula for the finite geometric series. If I've made a mistake, please let me know.)

Hi Pizza,

This is the sum of a GP  (Geometric progression) which maybe you have not learned about yet.

The first term, a=1       (3^0=1)

The common ratio r=3       (because you keep multiplying by 3)

there are n+1 terms here (see if you can understand why it is not n)

The formula for the sum of a GP is       $S_n=\frac{a(r^{n}-1)}{r-1}$

But in this case n needs to be replaced with n+1  because there are n+1 terms.

$S_{n+1}=\frac{a(r^{n+1}-1)}{r-1}\\ S_{n+1}=\frac{1(3^{n+1}-1)}{3-1}\\ S_{n+1}=\frac{(3^{n+1}-1)}{2}\\$

You seem to have a + where there should be a -