c $$ \sum_{r=1}^{n} 3 \cdot 2^{r-1}=381 $$

Question

UpStudy AI · Accepted Answer

To solve the equation:

$$
\sum_{r=1}^{n} 3 \cdot 2^{r-1} = 381
$$

Let's break it down step by step.

### Step 1: Understand the Summation
The summation can be expressed as:

$$
3 \cdot (2^0 + 2^1 + 2^2 + \dots + 2^{n-1})
$$

This is a geometric series where:
- **First term (a)** = $$2^0 = 1$$
- **Common ratio (r)** = 2
- **Number of terms** = $$n$$

### Step 2: Use the Formula for the Sum of a Geometric Series
The sum $$S$$ of the first $$n$$ terms of a geometric series is:

$$
S = a \cdot \frac{r^n - 1}{r - 1}
$$

Plugging in the values:

$$
2^0 + 2^1 + 2^2 + \dots + 2^{n-1} = \frac{2^n - 1}{2 - 1} = 2^n - 1
$$

### Step 3: Apply the Sum to the Original Equation
Substitute the sum back into the original equation:

$$
3 \cdot (2^n - 1) = 381
$$

Divide both sides by 3:

$$
2^n - 1 = 127
$$

Add 1 to both sides:

$$
2^n = 128
$$

### Step 4: Solve for $$n$$
Recognize that $$128$$ is a power of 2:

$$
2^7 = 128
$$

Therefore,

$$
n = 7
$$

### **Answer**
The value of $$ n $$ that satisfies the equation is **7**.
$$ n = 7 $$

c \( \sum_{r=1}^{n} 3 \cdot 2^{r-1}=381 \)