The nth term of a series is `3^n + 2n` ; find the sum of first n terms of the series.

Explanation

To find the sum of the first n terms of the series where the nth term is given by \( a_n = 3^n + 2n \), we can follow these steps:

### Step 1: Write the sum of the first n terms
The sum of the first n terms, denoted as \( S_n \), can be expressed as:
\[ S_n = a_1 + a_2 + a_3 + \ldots + a_n \]
Substituting the expression for \( a_n \):
\[ S_n = (3^1 + 2 \cdot 1) + (3^2 + 2 \cdot 2) + (3^3 + 2 \cdot 3) + \ldots + (3^n + 2n) \]

### Step 2: Separate the terms
We can separate the sum into two parts:
\[ S_n = \sum_{k=1}^{n} 3^k + \sum_{k=1}^{n} 2k \]
This simplifies to:
\[ S_n = \sum_{k=1}^{n} 3^k + 2 \sum_{k=1}^{n} k \]

### Step 3: Calculate the sum of the geometric series
The first part, \( \sum_{k=1}^{n} 3^k \), is a geometric series with the first term \( a = 3 \) and common ratio \( r = 3 \). The formula for the sum of the first n terms of a geometric series is:
\[ S_n = a \frac{r^n - 1}{r - 1} \]
Applying this formula:
\[ \sum_{k=1}^{n} 3^k = 3 \frac{3^n - 1}{3 - 1} = \frac{3(3^n - 1)}{2} \]

### Step 4: Calculate the sum of the first n natural numbers
The second part, \( \sum_{k=1}^{n} k \), is the sum of the first n natural numbers, which is given by:
\[ \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \]
Thus, we have:
\[ 2 \sum_{k=1}^{n} k = 2 \cdot \frac{n(n + 1)}{2} = n(n + 1) \]

### Step 5: Combine the results
Now, substituting back into the expression for \( S_n \):
\[ S_n = \frac{3(3^n - 1)}{2} + n(n + 1) \]

### Step 6: Simplify the expression
This gives us the final expression for the sum of the first n terms:
\[ S_n = \frac{3(3^n - 1)}{2} + n(n + 1) \]

### Final Answer
Thus, the sum of the first n terms of the series is:
\[ S_n = \frac{3(3^n - 1)}{2} + n(n + 1) \]