What geomatric series formula is used

Question

trying to follow a solution related to geometric series, but not sure what formula is used here. Any pointer is appreciated.Image here, can't embed image yet. I do try to plug in the Sn formula, but didn't get it to work

Answer 1

Notice:

$$3^n + 2(3^{n-1}) + 2^2 (3^{n-2}) + ... 2^{n-1}(3) + 2^n$$

can be effectively rewritten as

$$\sum_{k=0}^n 3^{n-k} \cdot 2^k = \sum_{k=0}^n 3^{n} \cdot 3^{-k} \cdot 2^k =\sum_{k=0}^n 3^{n} \cdot \frac{2^k}{3^k} = \sum_{k=0}^n 3^n \cdot \left( \frac{2}{3} \right)^k$$

This is a finite geometric series, then, with ratio $r=2/3$ and first term $2^03^n = 3^n$. The sum of a finite geometric series $a + ar + ar^2 + ... +ar^n$ is given by

$$\sum_{k=0}^n ar^k = \frac{a(r^{n+1} - 1)}{r-1}$$

Thus,

$$\sum_{k=0}^n 3^n \cdot \left( \frac{2}{3} \right)^k = \frac{3^n((2/3)^{n+1} - 1)}{2/3-1}$$

This is essentially the same expression as presented to you, just needing some manipulations to make the equivalence clear.

Multiply by $3$ on the top and bottom:

$$\frac{3^n((2/3)^{n+1} - 1)}{2/3-1} = \frac{3^{n+1} \cdot ((2/3)^{n+1} - 1)}{2 - 3}$$

Multiply by $-1$ on the top and bottom. We distribute this into the $()$ on top to reverse the order.

$$\frac{3^{n+1} \cdot ((2/3)^{n+1} - 1)}{2 - 3} = \frac{3^{n+1} \cdot (1 - (2/3)^{n+1})}{3 - 2}$$

Next:

$$1 - (2/3)^{n+1} = \frac{3^{n+1}}{3^{n+1}} - \frac{2^{n+1}}{3^{n+1}} = \frac{3^{n+1} - 2^{n+1}}{3^{n+1}}$$

The denominator cancels owing to the outside $3^{n+1}$ term, and of course $3-2=1$, leaving us with

$$\frac{3^{n+1} \cdot (1 - (2/3)^{n+1})}{3 - 2} = \frac{3^{n+1} \cdot \frac{3^{n+1} - 2^{n+1}}{3^{n+1}}}{1} = 3^{n+1} - 2^{n+1}$$

Thus,

$$3^n + 2(3^{n-1}) + ... +2^{n-1}(3) + 2^n = \sum_{k=0}^n 3^n \cdot \left( \frac{2}{3} \right)^k = 3^{n+1} - 2^{n+1}$$