Calculate the sum of each of the following geometric series: 1) $$ 6+12+24+\ldots+6144 $$ (2) $$ 4-8+16-\ldots+16384 $$ $$ -5+15-45+\ldots-3645 $$ (4) $$ 100-10+1-\ldots+\frac{1}{10000} $$ $$ 200\left(1+(1,01)+(1.01)^{2}+\ldots+(1,01)^{2 R}\right) $$ (6) $$ -12-6-3-\ldots-\frac{3}{256} $$

Question

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To calculate the sum of each geometric series, we will use the formula for the sum of a geometric series:

$$
S_n = a \frac{1 - r^n}{1 - r}
$$

where:
- $$ S_n $$ is the sum of the first $$ n $$ terms,
- $$ a $$ is the first term,
- $$ r $$ is the common ratio,
- $$ n $$ is the number of terms.

Let's analyze each series one by one.

### 1) $$ 6 + 12 + 24 + \ldots + 6144 $$

- First term $$ a = 6 $$
- Common ratio $$ r = \frac{12}{6} = 2 $$
- Last term $$ 6144 $$

To find $$ n $$, we can use the formula for the $$ n $$-th term of a geometric series:

$$
a_n = a \cdot r^{n-1}
$$

Setting $$ a_n = 6144 $$:

$$
6144 = 6 \cdot 2^{n-1}
$$

Dividing both sides by 6:

$$
1024 = 2^{n-1}
$$

Since $$ 1024 = 2^{10} $$, we have:

$$
n - 1 = 10 \implies n = 11
$$

Now we can calculate the sum:

$$
S_{11} = 6 \frac{1 - 2^{11}}{1 - 2} = 6 \frac{1 - 2048}{-1} = 6 \cdot 2047 = 12282
$$

### 2) $$ 4 - 8 + 16 - \ldots + 16384 $$

- First term $$ a = 4 $$
- Common ratio $$ r = -2 $$
- Last term $$ 16384 $$

Finding $$ n $$:

$$
16384 = 4 \cdot (-2)^{n-1}
$$

Dividing by 4:

$$
4096 = (-2)^{n-1}
$$

Since $$ 4096 = 2^{12} $$, we have:

$$
n - 1 = 12 \implies n = 13
$$

Now we can calculate the sum:

$$
S_{13} = 4 \frac{1 - (-2)^{13}}{1 - (-2)} = 4 \frac{1 + 8192}{3} = 4 \cdot \frac{8193}{3} = 4 \cdot 2731 = 10924
$$

### 3) $$ -5 + 15 - 45 + \ldots - 3645 $$

- First term $$ a = -5 $$
- Common ratio $$ r = -3 $$
- Last term $$ -3645 $$

Finding $$ n $$:

$$
-3645 = -5 \cdot (-3)^{n-1}
$$

Dividing by -5:

$$
729 = (-3)^{n-1}
$$

Since $$ 729 = 3^6 $$, we have:

$$
n - 1 = 6 \implies n = 7
$$

Now we can calculate the sum:

$$
S_{7} = -5 \frac{1 - (-3)^{7}}{1 - (-3)} = -5 \frac{1 + 2187}{4} = -5 \cdot \frac{2188}{4} = -5 \cdot 547 = -2735
$$

### 4) $$ 100 - 10 + 1 - \ldots + \frac{1}{10000} $$

- First term $$ a = 100 $$
- Common ratio $$ r = -0.1 $$
- Last term $$ \frac{1}{10000} $$

Finding $$ n $$:

$$
\frac{1}{10000} = 100 \cdot (-0.1)^{n-1}
$$

Dividing by 100:

$$
\frac{1}{100000} = (-0.1)^{n-1}
$$

Since $$ \frac{1}{100000} = (-0.1)^{5} $$, we have:

$$
n - 1 = 5 \implies n = 6
$$

Now we can calculate the sum:

$$
S_{6} = 100 \frac{1 - (-0.1)^{6}}{1 - (-0.1)} = 100 \frac{1 - \frac{1}{1000000}}{1.1} = 100 \cdot \frac{999999}{1000000 \cdot 1.1} = \frac{99999900}{1100000} \approx 90.909
$$

### 5) $$ 200\left(1 + (1.01) + (1.01)^{2} + \ldots + (1.01)^{2R}\right) $$

This is a geometric series with:

- First term $$ a = 1 $$
- Common ratio $$ r = 1.01 $$
- Number of terms $$ n = 2R + 1 $$

The sum is:

$$
S = 1 \frac{1 - (1.01)^{2R + 1}}{1 - 1.01} = \frac{1 - (1.01)^{2R + 1}}{-0.01} = -100(1 - (1.01)^{2R + 1})
$$

Thus, the total sum is:

$$
200 \cdot -100(1 - (1.01)^{2R + 1}) = -20000(1 - (1.01)^{2R + 1})
$$

### 6) $$ -12 - 6 - 3 - \ldots - \frac{3}{256} $$

- First term $$ a = -12 $$
- Common ratio $$ r = -\frac{1}{2} $$
- Last term $$ -\frac{3}{256} $$

Finding $$ n $$:

$$
-\frac{3}{256} = -12 \cdot \left(-\frac{1}{2}\right)^{n-1}
$$

Dividing by -12:

$$
\frac{1}{256} = \left(-\frac{1}{2}\right)^{n-1}
$$

Since $$ \frac{1}{256} = \left(\frac{1}{2}\right)^{8} $$, we have:

$$
n - 1 = 8 \implies n = 9
$$

Now we can calculate the sum:

\[
S_{9} = -12 \frac{1 - \left(-\frac{1}{2}\right)^{9}}{1 - \left(-\frac{1}{2}\right)} =
1. **First Series ($$6 + 12 + 24 + \ldots + 6144$$)**: - **Sum**: 12,282 2. **Second Series ($$4 - 8 + 16 - \ldots + 16384$$)**: - **Sum**: 10,924 3. **Third Series ($$-5 + 15 - 45 + \ldots - 3645$$)**: - **Sum**: -2,735 4. **Fourth Series ($$100 - 10 + 1 - \ldots + \frac{1}{10000}$$)**: - **Sum**: Approximately 90.909 5. **Fifth Series ($$200\left(1 + (1.01) + (1.01)^{2} + \ldots + (1.01)^{2R}\right)$$)**: - **Sum**: $$ -20,000(1 - (1.01)^{2R + 1}) $$ 6. **Sixth Series ($$-12 - 6 - 3 - \ldots - \frac{3}{256}$$)**: - **Sum**: $$-12 \frac{1 - \left(-\frac{1}{2}\right)^{9}}{1 - \left(-\frac{1}{2}\right)}$$

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