$$ \sum _ { k = 1 } ^ { m } 5 ( 3 ) ^ { k - 1 } = 65 $$

Question

UpStudy AI · Accepted Answer

To solve the equation:

$$
\sum_{k=1}^{m} 5 \cdot 3^{k-1} = 65
$$

we recognize that this is a geometric series. Let's break it down step by step.

### Step 1: Understand the Geometric Series

A geometric series has the form:

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

where:
- $$ S $$ is the sum of the series,
- $$ a $$ is the first term,
- $$ r $$ is the common ratio,
- $$ n $$ is the number of terms.

### Step 2: Identify the Components

In the given series:

$$
a = 5 \cdot 3^{1-1} = 5 \cdot 3^0 = 5 \
r = 3 \
S = 65 \
$$

### Step 3: Plug into the Geometric Series Formula

$$
65 = 5 \cdot \frac{3^m - 1}{3 - 1}
$$

Simplify the denominator:

$$
65 = 5 \cdot \frac{3^m - 1}{2}
$$

### Step 4: Solve for $$ 3^m $$

Multiply both sides by 2:

$$
130 = 5 \cdot (3^m - 1)
$$

Divide both sides by 5:

$$
26 = 3^m - 1
$$

Add 1 to both sides:

$$
27 = 3^m
$$

### Step 5: Find $$ m $$

Recognize that $$ 27 $$ is a power of $$ 3 $$:

$$
3^3 = 27
$$

Therefore:

$$
m = 3
$$

### Conclusion

The value of $$ m $$ that satisfies the equation is **3**.
$$ m = 3 $$

\( \sum _ { k = 1 } ^ { m } 5 ( 3 ) ^ { k - 1 } = 65 \)