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

Explanation

To evaluate the sum \( \Sigma_{n=1}^{6} (3 + 2^n) \), we can break it down step by step:

### Step 1: Write out the terms of the summation
We need to evaluate the expression for each integer \( n \) from 1 to 6.

\[ \Sigma_{n=1}^{6} (3 + 2^n) = (3 + 2^1) + (3 + 2^2) + (3 + 2^3) + (3 + 2^4) + (3 + 2^5) + (3 + 2^6) \]

### Step 2: Calculate each term
Now, we calculate each term individually:

- For \( n = 1 \): \( 3 + 2^1 = 3 + 2 = 5 \)
- For \( n = 2 \): \( 3 + 2^2 = 3 + 4 = 7 \)
- For \( n = 3 \): \( 3 + 2^3 = 3 + 8 = 11 \)
- For \( n = 4 \): \( 3 + 2^4 = 3 + 16 = 19 \)
- For \( n = 5 \): \( 3 + 2^5 = 3 + 32 = 35 \)
- For \( n = 6 \): \( 3 + 2^6 = 3 + 64 = 67 \)

### Step 3: Sum the calculated terms
Now, we sum all the calculated terms:

\[ 5 + 7 + 11 + 19 + 35 + 67 \]

Calculating this step by step:

- \( 5 + 7 = 12 \)
- \( 12 + 11 = 23 \)
- \( 23 + 19 = 42 \)
- \( 42 + 35 = 77 \)
- \( 77 + 67 = 144 \)

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

---