Why is $2 \cdot \sum_{j=0}^{i-1} 3^j = 3^i - 1$?

Question

I recentely encountered the following equation:

$2 \cdot \sum_{j=0}^{i-1} 3^j = 3^i - 1$

I've tried a few numbers and this seems to work. Could someone explain to my why this is the case or post a proof?

I'm not even sure how to properly tag or google this, so suggestions for that would also be great!

Answer 1

This is the sum of Geometric series

$a+ar+ar^2 +\cdots ar^{n-1} = a\frac{r^n-1}{r-1}$

Here we have,

$2(1+3 +3^2+\cdots3^{i-1})= 2 \frac{3^i-1}{3-1}=3^i-1$

Answer 2

In base 10, we have

$9=10-1$

$99=100-1$

$999=1000-1$

etc.

The result you've stated is simply that but in base 3, i.e.

$2_3=10_3-1_3$

$22_3=100_3-1_3$

$222_3=1000_3-1_3$

etc., where $x_3$ here means that $x$ is a number written in base $3$. Someone else has already posted how to prove the result.

Answer 3

Several answers and comments tell you how to do this with a geometric series.

For another approach, think about why $10^n-1 = 99\ldots 9$ and then think about the largest $n$-digit number you can write in base $3$.