Rewriting the nth term of a series in the form $ar^{n-1}$

Question

In the formula below, where does the $\frac{4}{3}$ come from and what happened to the $3$? How did they get the far right answer? Taken from Stewart Early Transcendentals Calculus textbook.

$$\sum^\infty_{n=1} 2^{2n}3^{1-n}=\sum^\infty_{n=1}(2^2)^{n}3^{-(n-1)}=\sum^\infty_{n=1}\frac{4^n}{3^{n-1}}=\sum_{n=1}^\infty4\left(\frac{4}{3}\right)^{n-1}$$

Answer 1

$$\frac{4^n}{3^{n-1}}=\frac{4^{1+(n-1)}}{3^{n-1}}$$

$$=\frac{4^1\cdot 4^{n-1}}{3^{n-1}}$$

$$=4\cdot \frac{4^{n-1}}{3^{n-1}}$$

$$=4\cdot\big(\frac{4}{3}\big)^{n-1}$$

Answer 2

$$\frac{4^n}{3^{n-1}}=\frac{4^1\cdot 4^{n-1}}{3^{n-1}}=4\cdot\frac {4^{n-1}}{3^{n-1}}=4\cdot\left(\frac43\right)^{n-1}$$

Answer 3

You have $$ 2^{2n}3^{1-n} = (2^2)^n3^{-(n-1)} = 4^n3^{-(n-1)} = \frac{4^n}{3^{n-1}} = \frac{4\cdot4^{n-1}}{3^{n-1}} = 4\left(\frac{4}{3}\right)^{n-1} $$ Here we have used the formulas

1. $$ a^ma^k = a^{m+k} $$
2. $$ a^{-m} = \frac{1}{a^m} $$
3. $$ (a^m)^k = a^{mk} $$
4. $$ \frac{a^m}{b^m} = \left(\frac{a}{b}\right)^{m} $$ So

In the formula below, where does the $\frac{4}{3}$ come from

It came from moving the exponent outside using formula 4 above.

and what happened to the 3?

Nothing happened to the $3$.

How did they get the far right answer?

The last step is using the formula 4 from above.