Question

find the sum of the of the following series round to the nearest hundreth if necessary
5+10+20+...+81920
sum of a finite geometric series
S(under)n=a(under)1-a(under)1r^n/1-r

Answer 1

To find the sum of the geometric series \( 5 + 10 + 20 + \ldots + 81920 \), we first need to identify the first term \( a \), the common ratio \( r \), and the number of terms \( n \).

### Step 1: Identify the terms

1. **First term (\( a \))**: The first term \( a = 5 \).
2. **Common ratio (\( r \))**: To find the common ratio, notice that each term is multiplied by \( 2 \) to get the next term:
\[
r = \frac{10}{5} = 2
\]
3. **Last term**: The last term we have is \( 81920 \).

### Step 2: Calculate the number of terms (\( n \))

The \( n \)-th term of a geometric series can be expressed as:
\[
a_n = a \cdot r^{n-1}
\]
Setting this equal to the last term \( 81920 \):
\[
5 \cdot 2^{n-1} = 81920
\]
Now, divide both sides by \( 5 \):
\[
2^{n-1} = \frac{81920}{5} = 16384
\]
Next, we can rewrite \( 16384 \) as a power of \( 2 \):
\[
16384 = 2^{14}
\]
So we have:
\[
2^{n-1} = 2^{14}
\]
Thus, equating the exponents gives:
\[
n - 1 = 14 \implies n = 15
\]

### Step 3: Sum the series

We can now use the formula for the sum of a finite geometric series:
\[
S_n = \frac{a(1 - r^n)}{1 - r}
\]
Substituting \( a = 5 \), \( r = 2 \), and \( n = 15 \):
\[
S_{15} = \frac{5(1 - 2^{15})}{1 - 2}
\]
Calculating \( 2^{15} \):
\[
2^{15} = 32768
\]
Now substitute this back into the sum formula:
\[
S_{15} = \frac{5(1 - 32768)}{-1} = \frac{5 \cdot (-32767)}{-1} = 5 \cdot 32767 = 163835
\]

### Final Answer
Thus, the sum of the series \( 5 + 10 + 20 + \ldots + 81920 \) is:
\[
\boxed{163835}
\]