Find the sum of the first n terms of the following geometric series: a) 2/27 + 2/9 + 2/3 + ... to 9 terms. b) -64 + 32 - 16 + ... to 10 terms.

The formula for the sum of the first $n$ terms of a geometric series is $S_n = \frac{a(r^n - 1)}{r-1}$ or $S_n = \frac{a(1 - r^n)}{1-r}$, where $a$ is the first term and $r$ is the common ratio.

1.a)

Step 1: Identify the first term ($a$), common ratio ($r$), and number of terms ($n$).
The series is $\frac{2}{27} + \frac{2}{9} + \frac{2}{3} + \dots$ to 9 terms.
$a = \frac{2}{27}$
$r = \frac{2/9}{2/27} = \frac{2}{9} \times \frac{27}{2} = 3$
$n = 9$

Step 2: Apply the sum formula.

$$S_9 = \frac{a(r^n - 1)}{r-1}$$ $$S_9 = \frac{\frac{2}{27}(3^9 - 1)}{3-1}$$ $$S_9 = \frac{\frac{2}{27}(19683 - 1)}{2}$$ $$S_9 = \frac{\frac{2}{27}(19682)}{2}$$ $$S_9 = \frac{19682}{27}$$

The sum of the series is $\boxed{\frac{19682}{27}}$.

1.b)

Step 1: Identify the first term ($a$), common ratio ($r$), and number of terms ($n$).
The series is $-64 + 32 - 16 + \dots$ to 10 terms.
$a = -64$
$r = \frac{32}{-64} = -\frac{1}{2}$
$n = 10$

Step 2: Apply the sum formula.
Since $|r| < 1$, we use $S_n = \frac{a(1 - r^n)}{1-r}$.

$$S_{10} = \frac{-64\left(1 - \left(-\frac{1}{2}\right)^{10}\right)}{1 - \left(-\frac{1}{2}\right)}$$ $$S_{10} = \frac{-64\left(1 - \frac{1}{1024}\right)}{1 + \frac{1}{2}}$$ $$S_{10} = \frac{-64\left(\frac{1024 - 1}{1024}\right)}{\frac{3}{2}}$$ $$S_{10} = \frac{-64\left(\frac{1023}{1024}\right)}{\frac{3}{2}}$$ $$S_{10} = -64 \times \frac{1023}{1024} \times \frac{2}{3}$$ $$S_{10} = -\frac{1023 \times 2}{16 \times 3}$$ $$S_{10} = -\frac{1023}{8 \times 3}$$ $$S_{10} = -\frac{341}{8}$$

The sum of the series is $\boxed{-\frac{341}{8}}$.

2)

Step 1: Identify the first term ($a$), common ratio ($r$), and number of terms ($n$).
The sequence is $\frac{1}{4}, -\frac{1}{2}, 1, \dots$ and we need the sum of the first 12 terms.
$a = \frac{1}{4}$
$r = \frac{-1/2}{1/4} = -\frac{1}{2} \times 4 = -2$
$n = 12$

Step 2: Apply the sum formula.

$$S_{12} = \frac{a(r^n - 1)}{r-1}$$ $$S_{12} = \frac{\frac{1}{4}((-2)^{12} - 1)}{-2-1}$$ $$S_{12} = \frac{\frac{1}{4}(4096 - 1)}{-3}$$ $$S_{12} = \frac{\frac{1}{4}(4095)}{-3}$$ $$S_{12} = \frac{4095}{-12}$$ $$S_{12} = -\frac{1365}{4}$$

The sum of the first 12 terms is $\boxed{-\frac{1365}{4}}$.

3.a)

Step 1: Identify the first term ($a$), common ratio ($r$), and number of terms ($n$) from the summation.
The summation is $\sum_{K=1}^{10} \frac{1}{50}(5)^{K-1}$.
This is a geometric series.
For $K=1$, the first term $a = \frac{1}{50}(5)^{1-1} = \frac{1}{50}(5)^0 = \frac{1}{50} \times 1 = \frac{1}{50}$.
The common ratio $r = 5$.
The number of terms $n = 10 - 1 + 1 = 10$.

Step 2: Apply the sum formula.

$$S_{10} = \frac{a(r^n - 1)}{r-1}$$ $$S_{10} = \frac{\frac{1}{50}(5^{10} - 1)}{5-1}$$ $$S_{10} = \frac{\frac{1}{50}(5^{10} - 1)}{4}$$ $$S_{10} = \frac{5^{10} - 1}{200}$$ $$S_{10} = \frac{9765625 - 1}{200}$$ $$S_{10} = \frac{9765624}{200}$$ $$S_{10} = 48828.12$$

The sum is $\boxed{48828.12}$.

3.b)

Step 1: Identify the first term ($a$), common ratio ($r$), and number of terms ($n$) from the summation.
The summation is $\sum_{i=0}^{10} 3^{4-i}$.
This is a geometric series.
For $i=0$, the first term $a = 3^{4-0} = 3^4 = 81$.
For $i=1$, the second term is $3^{4-1} = 3^3 = 27$.
The common ratio $r = \frac{27}{81} = \frac{1}{3}$.
The number of terms $n = 10 - 0 + 1 = 11$.

Step 2: Apply the sum formula.
Since $|r| < 1$, we use $S_n = \frac{a(1 - r^n)}{1-r}$.

$$S_{11} = \frac{81\left(1 - \left(\frac{1}{3}\right)^{11}\right)}{1 - \frac{1}{3}}$$ $$S_{11} = \frac{81\left(1 - \frac{1}{3^{11}}\right)}{\frac{2}{3}}$$ $$$$