`"Evaluate "Sigma_(n=1)^(6) (3+2^(n)).`

Explanation

To evaluate the summation \( \Sigma_{n=1}^{6} (3 + 2^n) \), we can break it down into simpler parts.

### Step-by-Step Solution:

1. **Expand the Summation**:
We start by expanding the summation:
\[ \Sigma_{n=1}^{6} (3 + 2^n) = (3 + 2^1) + (3 + 2^2) + (3 + 2^3) + (3 + 2^4) + (3 + 2^5) + (3 + 2^6) \]

2. **Separate the Terms**:
We can separate the constant term from the variable term:
\[ = \Sigma_{n=1}^{6} 3 + \Sigma_{n=1}^{6} 2^n \]

3. **Calculate the First Summation**:
The first summation is simply:
\[ \Sigma_{n=1}^{6} 3 = 3 \times 6 = 18 \]

4. **Calculate the Second Summation**:
The second summation \( \Sigma_{n=1}^{6} 2^n \) is a geometric series where:
- First term \( a = 2^1 = 2 \)
- Common ratio \( r = 2 \)
- Number of terms \( n = 6 \)

The formula for the sum of the first \( n \) terms of a geometric series is:
\[ S_n = a \frac{r^n - 1}{r - 1} \]

Plugging in the values:
\[ S_6 = 2 \frac{2^6 - 1}{2 - 1} = 2 \frac{64 - 1}{1} = 2 \times 63 = 126 \]

5. **Combine the Results**:
Now, we combine both parts:
\[ \Sigma_{n=1}^{6} (3 + 2^n) = 18 + 126 = 144 \]

### Final Answer:
Thus, the value of the summation \( \Sigma_{n=1}^{6} (3 + 2^n) \) is \( 144 \).

---