`"Evaluate Sigma_(n=1)^(11) (2+3^(n))`

Explanation

To evaluate the sum \( \Sigma_{n=1}^{11} (2 + 3^n) \), we can break it down into two separate summations.

### Step-by-Step Solution:

1. **Separate the Summation**:
\[ \Sigma_{n=1}^{11} (2 + 3^n) = \Sigma_{n=1}^{11} 2 + \Sigma_{n=1}^{11} 3^n \]

2. **Calculate the First Summation**:
The first summation is a constant:
\[ \Sigma_{n=1}^{11} 2 = 2 \times 11 = 22 \]

3. **Calculate the Second Summation**:
The second summation is a geometric series:
\[ \Sigma_{n=1}^{11} 3^n \]
The formula for the sum of the first \( n \) terms of a geometric series is:
\[ S_n = a \frac{r^n - 1}{r - 1} \]
where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms. Here, \( a = 3 \), \( r = 3 \), and \( n = 11 \).

Plugging in the values:
\[ S_{11} = 3 \frac{3^{11} - 1}{3 - 1} = 3 \frac{3^{11} - 1}{2} \]

4. **Calculate \( 3^{11} \)**:
First, we need to calculate \( 3^{11} \):
\[ 3^{11} = 177147 \]
Therefore,
\[ S_{11} = 3 \frac{177147 - 1}{2} = 3 \frac{177146}{2} = 3 \times 88573 = 265719 \]

5. **Combine the Results**:
Now, we combine both parts of the summation:
\[ \Sigma_{n=1}^{11} (2 + 3^n) = 22 + 265719 = 265741 \]

### Final Answer:
\[ \Sigma_{n=1}^{11} (2 + 3^n) = 265741 \]