1. **State the problem:** We want to write the sum $3 + 6 + 12 + \dots + 192$ using sigma notation and then evaluate it. 2. **Identify the sequence type:** This is a geometric sequence where the first term $a_1 = 3$ and the common ratio $r = \frac{6}{3} = 2$. 3. **Write the sum in sigma notation:** The $n$th term of a geometric sequence is $a_n = a_1 r^{n-1}$. So the sum can be written as: $$\sum_{n=1}^k 3 \cdot 2^{n-1}$$ where $k$ is the number of terms. 4. **Find the number of terms $k$:** The last term is 192, so: $$3 \cdot 2^{k-1} = 192$$ Divide both sides by 3: $$\cancel{3} \cdot 2^{k-1} = \frac{192}{\cancel{3}}$$ $$2^{k-1} = 64$$ Since $64 = 2^6$, we have: $$2^{k-1} = 2^6 \implies k-1 = 6 \implies k = 7$$ 5. **Use the sum formula for geometric series:** $$S_k = \frac{a_1 (1 - r^k)}{1 - r}$$ Substitute $a_1 = 3$, $r = 2$, and $k = 7$: $$S_7 = \frac{3 (1 - 2^7)}{1 - 2} = \frac{3 (1 - 128)}{1 - 2} = \frac{3 (-127)}{-1}$$ 6. **Simplify the sum:** $$S_7 = \frac{3 \times (-127)}{-1} = 381$$ **Final answer:** $$\sum_{n=1}^7 3 \cdot 2^{n-1} = 381$$