Evaluating $\sum\limits_{n=0}^{20} \frac{(-1)^{n}2^{n+1}}{3^{n}},$ I know the sum of the series $$2 - \frac{4}{3} + \frac{8}{9} - \cdots + \frac{(-1)^{20}2^{21}}{3^{20}}$$ is equal to $$\sum\limits_{n=0}^{20} \frac{(-1)^{n}2^{n+1}}{3^{n}},$$ but I don't know how to calculate the sum without manually entering it into the calculator.

$$\sum_{n=1}^{20}\frac{(-1)^n2^{n+1}}{3^n}=2\sum_{n=1}^{20}\left(-\frac{2}{3}\right)^n=2\left(-\frac{2}{3}\right)\frac{\left(-\frac{2}{3}\right)^{20}-1}{\left(-\frac{2}{3}-1\right)}=$$
$$=\left(-\frac{4}{3}\right)\left(-\frac{3}{5}\right)\frac{2^{20}-3^{20}}{3^{20}}=\frac{4}{5}\frac{2^{20}-3^{20}}{3^{20}}$$
