Question

find the sum of the of the following series round to the nearest hundreth if necessary
5++15+45...+885735
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 given geometric series \(5 + 15 + 45 + \ldots + 885735\), we first need to identify the initial term \(a\), the common ratio \(r\), and the number of terms \(n\).

1. **Identify the terms**:
- The first term \(a_1 = 5\).
- The second term is \(15\), which is \(3\) times the first term: \(15 = 5 \cdot 3\).
- The third term is \(45\), which is again \(3\) times the second term: \(45 = 15 \cdot 3\).

Hence, the common ratio \(r\) is \(3\).

2. **Identify the last term**: We know the last term \(a_n = 885735\).

3. **Find the number of terms \(n\)**:
We know for a geometric series:
\[
a_n = a \cdot r^{n-1}
\]
So:
\[
885735 = 5 \cdot 3^{n-1}
\]

To find \(n\), we can first divide both sides by \(5\):
\[
177147 = 3^{n-1}
\]

Next, we can express \(177147\) as a power of \(3\):
- Calculating powers of \(3\),
- \(3^1 = 3\)
- \(3^2 = 9\)
- \(3^3 = 27\)
- \(3^4 = 81\)
- \(3^5 = 243\)
- \(3^6 = 729\)
- \(3^7 = 2187\)
- \(3^8 = 6561\)
- \(3^9 = 19683\)
- \(3^{10} = 59049\)
- \(3^{11} = 177147\)

Since \(177147 = 3^{11}\), we can set:
\[
n - 1 = 11 \implies n = 12
\]

4. **Calculate the sum**:
Using the formula for the sum \(S_n\) of a finite geometric series:
\[
S_n = a \cdot \frac{1 - r^n}{1 - r}
\]
where:
- \(a = 5\),
- \(r = 3\),
- \(n = 12\).

Therefore:
\[
S_{12} = 5 \cdot \frac{1 - 3^{12}}{1 - 3}
\]
Calculate \(3^{12}\):
\[
3^{12} = 531441
\]

Now substitute:
\[
S_{12} = 5 \cdot \frac{1 - 531441}{-2} = 5 \cdot \frac{-531440}{-2} = 5 \cdot 265720 = 1328600
\]

Thus, the sum of the series \(5 + 15 + 45 + \ldots + 885735\) is **1328600**.