To find the sum to \( n \) terms of the series \( 5, 7, 13, 31, 85, \ldots \), we will follow these steps:
### Step 1: Identify the pattern in the series
Let's denote the terms of the series as follows:
- \( a_1 = 5 \)
- \( a_2 = 7 \)
- \( a_3 = 13 \)
- \( a_4 = 31 \)
- \( a_5 = 85 \)
We can observe how each term relates to the previous ones.
### Step 2: Calculate the differences between consecutive terms
Let's calculate the differences between consecutive terms:
- \( a_2 - a_1 = 7 - 5 = 2 \)
- \( a_3 - a_2 = 13 - 7 = 6 \)
- \( a_4 - a_3 = 31 - 13 = 18 \)
- \( a_5 - a_4 = 85 - 31 = 54 \)
Now, let's write these differences:
- \( 2, 6, 18, 54 \)
### Step 3: Identify the pattern in the differences
Next, we can observe the ratios of these differences:
- \( \frac{6}{2} = 3 \)
- \( \frac{18}{6} = 3 \)
- \( \frac{54}{18} = 3 \)
This indicates that the differences are in a geometric progression with a common ratio of \( 3 \).
### Step 4: Express the \( n \)-th term
The first difference is \( 2 \) (which is \( 3^1 \)), and the subsequent differences can be expressed as:
- \( 2 \cdot 3^{n-1} \) for the \( n \)-th term.
Thus, we can express the \( n \)-th term \( a_n \) as:
\[
a_n = 5 + \sum_{k=1}^{n-1} 2 \cdot 3^{k-1}
\]
### Step 5: Calculate the sum of the geometric series
The sum of the geometric series can be calculated using the formula:
\[
S = a \frac{r^n - 1}{r - 1}
\]
where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms.
In our case:
- First term \( a = 2 \)
- Common ratio \( r = 3 \)
- Number of terms \( n-1 \)
Thus, the sum becomes:
\[
\sum_{k=1}^{n-1} 2 \cdot 3^{k-1} = 2 \cdot \frac{3^{n-1} - 1}{3 - 1} = 2 \cdot \frac{3^{n-1} - 1}{2} = 3^{n-1} - 1
\]
### Step 6: Write the \( n \)-th term
Now substituting back into the expression for \( a_n \):
\[
a_n = 5 + (3^{n-1} - 1) = 3^{n-1} + 4
\]
### Step 7: Find the sum of the first \( n \) terms
To find the sum of the first \( n \) terms \( S_n \):
\[
S_n = \sum_{k=1}^{n} a_k = \sum_{k=1}^{n} (3^{k-1} + 4)
\]
This can be split into two sums:
\[
S_n = \sum_{k=1}^{n} 3^{k-1} + \sum_{k=1}^{n} 4
\]
The first sum is again a geometric series:
\[
\sum_{k=1}^{n} 3^{k-1} = \frac{3^n - 1}{3 - 1} = \frac{3^n - 1}{2}
\]
The second sum is simply:
\[
\sum_{k=1}^{n} 4 = 4n
\]
### Final expression for \( S_n \)
Combining these results gives:
\[
S_n = \frac{3^n - 1}{2} + 4n
\]
### Conclusion
Thus, the sum to \( n \) terms of the series \( 5, 7, 13, 31, 85, \ldots \) is:
\[
S_n = \frac{3^n - 1}{2} + 4n
\]
