Rewriting the nth term of a series in the form $ar^{n-1}$ 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.
$$\sum^\infty_{n=1} 2^{2n}3^{1-n}=\sum^\infty_{n=1}(2^2)^{n}3^{-(n-1)}=\sum^\infty_{n=1}\frac{4^n}{3^{n-1}}=\sum_{n=1}^\infty4\left(\frac{4}{3}\right)^{n-1}$$

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